Okay, let's tackle this problem. The question is asking for the radius of a sphere where its volume and surface area are numerically equal. That means the number representing the volume (in cubic units) is the same as the number representing the surface area (in square units). So, even though their units are different (cubic vs. square), their numerical values are the same. 

First, I need to recall the formulas for the volume and surface area of a sphere. The volume of a sphere is given by (4/3)πr³, and the surface area is 4πr². Since the problem states that these two values are numerically equal, I can set them equal to each other and solve for r.

So, setting the formulas equal:

(4/3)πr³ = 4πr²

Hmm, let's see. Both sides have 4π, so maybe I can divide both sides by 4π to simplify. Let me do that:

( (4/3)πr³ ) / (4π) = (4πr²) / (4π)

Simplifying the left side: the 4π cancels out, leaving (1/3)r³.

On the right side: 4π cancels out, leaving r².

So now we have:

(1/3)r³ = r²

Hmm, that's simpler. Now, let's solve for r. Let's subtract r² from both sides to bring all terms to one side:

(1/3)r³ - r² = 0

Factor out an r² from both terms:

r²( (1/3)r - 1 ) = 0

So, either r² = 0 or (1/3)r - 1 = 0.

But r² = 0 would imply r = 0, which doesn't make sense for a sphere. The radius can't be zero because then it wouldn't be a sphere. So we discard that solution.

Therefore, we solve (1/3)r - 1 = 0:

(1/3)r = 1

Multiply both sides by 3:

r = 3

So, the radius should be 3 units. Let me check if that works.

Calculating the volume: (4/3)π(3)³ = (4/3)π(27) = 36π

Surface area: 4π(3)² = 4π(9) = 36π

Yes, both are 36π, so their numerical values are equal. Therefore, the radius is 3 units. That seems right. 

Wait, just to make sure I didn't make any algebraic mistakes. Let me go through the steps again.

Starting with (4/3)πr³ = 4πr². Dividing both sides by 4π gives (1/3)r³ = r². Then moving all terms to one side: (1/3)r³ - r² = 0. Factoring r² gives r²( (1/3)r - 1 ) = 0. So solutions are r = 0 or r = 3. Correct. So 3 is the valid solution. 

Alternatively, maybe if I divide both sides by r² (assuming r ≠ 0), then (4/3)πr = 4π. Then cancel 4π from both sides: (1/3)r = 1, so r = 3. Either way, same result. 

Therefore, the radius is 3 units. I think that's solid. No mistakes here. The answer should be 3.
Therefore, the radius of the sphere is $\boxed{3}$ units.
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