Solution:

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 $$\left\lfloor \frac{216}{32\pi/3}\right\rfloor=\left\lfloor\frac{81}{4\pi}\right\rfloor.$$

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.