2021 Fall AMC 10A Problem 3

The cube has volume $6^3=216.$ One ball of clay has volume $\frac{4}{3}\pi\cdot 2^3=\frac{32\pi}{3}.$ Because the clay may be reshaped but not compressed, the maximum number of balls is $$\left\lfloor \frac{216}{\frac{32\pi}{3}}\right\rfloor=\left\lfloor\frac{81}{4\pi}\right\rfloor.$$ Since $12\lt 4\pi\lt 13,$ we have $6\lt \frac{81}{4\pi}\lt \frac{81}{12}\lt 7.$ Therefore the floor is $6.$ Thus, D is the correct answer.