Well, wouldn't it be $100$ balls? If the radius is $1/2$, then the ball's diameter would be equal to $1$. Thus, the number of balls that can fit into the box with the given dimensions is equal to $ (10/1)\times\ (10/1)\times\ (1/1) = 100$

That's actually not the optimal packing. You can fit $106$ balls by packing rows of 10,9,10,9,10,9,10,9,10,10,10. I think that's optimal, although I don't have a proof.

Yeah, i just actually realized that there was a better way of packing the balls. I think I can prove that i can fit $106$ circles of radius 1/2 into a 10 by 10 rectangle. I'm trying to use the same logic though now to apply it to 3-D...might take some time.

You don't have to do any work to extend it to three dimensions, at least for this problem. Since the box has a thickness of 1 unit, all the balls are fixed along that dimension, and taking the cross-section with a plane 1/2 unit along this dimension yields circles of radius 1/2 in a 10x10 box. So the problem is exactly the same as the 2-D 10x10 problem.