imc

The volume of the cube is $V_{\text{cube}}=6^3=216,$ and the volume of a clay ball is $V_{\text{ball}}=\frac{4}{3}\cdot \pi \cdot 2^3=\frac{32}{3}\pi .$ Since the balls can be reshaped but not compressed, the maximum number of balls that can completely fit inside a cube is $$\lfloor \frac{V_{\text{cube}}}{V_{\text{ball}}}\rfloor =\lfloor \frac{81}{4\pi }\rfloor .$$ Approximating with $\pi \approx 3.14,$ we have $12<4\pi <13,$ or $\lfloor \frac{81}{13}\rfloor \le \lfloor \frac{81}{4\pi }\rfloor \le \lfloor \frac{81}{12}\rfloor .$ We simplify to get $$6\le \lfloor \frac{81}{4\pi }\rfloor \le 6,$$ from which $\lfloor \frac{81}{4\pi }\rfloor =\boxed{6}.$