<Question>
What is the length, in units, of the radius of a sphere whose volume and surface area, in cubic units and square units, respectively, are numerically equal?
</Question>

<think>
<Plan_1>
Okay, so I need to find the radius of a sphere where the volume and the surface area are numerically equal. That means, even though volume is in cubic units and surface area is in square units, their numerical values are the same. So, if the radius is r, then the volume is (4/3)πr³ and the surface area is 4πr². I need to set these two expressions equal to each other and solve for r.

Let me write that down:
(4/3)πr³ = 4πr²

Hmm, okay. Let me see. First, I can simplify this equation. Both sides have 4πr². Maybe I can divide both sides by 4πr² to simplify.
Let me try that.

<prompt_1.1> Simplify the equation (4/3)πr³ = 4πr² by dividing both sides by 4πr². </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
Dividing both sides by 4πr² gives:

( (4/3)πr³ ) / (4πr²) ) = (4πr²) / (4πr²)

Simplifying the left side: the 4 cancels with 4, π cancels with π, r³ divided by r² is r. So left side is (1/3)r. The right side is 1. So now the equation is:

(1/3)r = 1
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
To solve for r, multiply both sides by 3:

<prompt_2.1> Solve the equation (1/3)r = 1 for r by multiplying both sides by 3. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
r = 3

Wait, so is the radius 3 units?

Let me check if that makes sense.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
If r = 3, then let's compute the volume and surface area.

<prompt_3.1> Calculate the volume and surface area when r = 3 and verify they are numerically equal. </prompt_3.1>

Hmm, that seems straightforward. Did I miss something here? Let me check my steps again.

<prompt_3.2> Recheck the simplification process of the equation (4/3)πr³ = 4πr². </prompt_3.2>

But let me confirm with another approach. Let's suppose I set up the equation again:

<prompt_3.3> Verify the solution by setting up the equation (4/3)πr³ = 4πr² again and solving for r. </prompt_3.3>

Alternatively, maybe there's another solution? Let me see.

<prompt_3.4> Explore if there are any other solutions to the equation (4/3)πr³ = 4πr². </prompt_3.4>

Wait, but maybe I should consider if there's a case where r=0?

<prompt_3.5> Consider the case where r = 0 for the equation (4/3)πr³ = 4πr². </prompt_3.5>

Another way to look at this is to consider the general equation for volume and surface area being equal.

<prompt_3.6> Discuss the general solution for the equation (4/3)πr³ = 4πr² and confirm the solution r = 3. </prompt_3.6>
</Plan_3>

<Execution_3>
<execution_3.1>
Volume: (4/3)π(3)³ = (4/3)π*27 = (4/3)*27*π = 36π

Surface area: 4π(3)² = 4π*9 = 36π

Oh, wait! They are both 36π. So the numerical values are equal when r = 3. So even though the units are different (cubic vs. square), their numerical values are the same.
</execution_3.1>

<execution_3.2>
Original equation: (4/3)πr³ = 4πr²

Divide both sides by 4πr²: (1/3)r = 1

Multiply both sides by 3: r = 3
</execution_3.2>

<execution_3.3>
Volume = (4/3)π(3)³ = (4/3)π*27 = (4/3)*27*π = 36π

Surface area = 4π(3)² = 4π*9 = 36π
</execution_3.3>

<execution_3.4>
Original equation:
(4/3)πr³ = 4πr²

Divide both sides by 4π:
(1/3)r³ = r²

Multiply both sides by 3:
r³ = 3r²

Subtract 3r² from both sides:
r³ - 3r² = 0

Factor out r²:
r²(r - 3) = 0

So, the solutions are r² = 0 => r = 0, or r - 3 = 0 => r = 3. Since a sphere with radius 0 doesn't make sense in this context, the only valid solution is r = 3.
</execution_3.4>

<execution_3.5>
But if r=0, both volume and surface area would be zero, which is technically equal, but a sphere with radius 0 isn't meaningful in this context. So we can discard r=0 as a trivial solution. Therefore, the only valid solution is r=3.

But just to make sure, let's plug r=3 back into the original formulas:

Volume: (4/3)π(3)³ = (4/3)π(27) = (4/3)(27)π = 36π

Surface area: 4π(3)² = 4π(9) = 36π
</execution_3.5>

<execution_3.6>
So, if I set (4/3)πr³ = 4πr², then dividing both sides by 4πr² (assuming r ≠ 0), we get (1/3)r = 1, so r=3. If r=0, then both volume and surface area would be zero, but a sphere with radius 0 isn't really a sphere. So, the only valid solution is r=3.
</execution_3.6>
</Execution_3>

<Plan_4>
Based on execution_3.1, execution_3.2, execution_3.3, execution_3.4, execution_3.5, execution_3.6:
Therefore, the radius is 3 units.

I think that's thorough. Let me check if there's another way to approach this problem. Maybe using ratios? The ratio of volume to surface area for a sphere is (4/3)πr³ divided by 4πr², which simplifies to (1/3)r.

<prompt_4.1> Use the ratio of volume to surface area to confirm the solution for the radius. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
So if the numerical values are equal, then (1/3)r = 1, so r=3. Yep, same result. That's another way to see it. The ratio of volume to surface area for a sphere is (1/3)r, so setting that equal to 1 gives r=3.
</execution_4.1>
</Execution_4>

<Final_answer>
Therefore, the length of the radius of the sphere is $\boxed{3}$ units.
</Final_answer>
</think>