Volume of a Cube is Side $$^3=125 \Rightarrow$$ Side $$=\sqrt[3]{125}=5$$

The minimum possible height of the cylinder, in which a cube of edge 5 each is inscribed, can be 5 only and the diagonal of the one of the faces of cube i.e. $$\sqrt{2} \times \operatorname{Side}=\sqrt{2} \times 5=5 \sqrt{2}$$ is equal to the diameter of the cylinder, so we get $$2 \mathrm{r}=5 \sqrt{2} \Rightarrow \mathrm{r}=\frac{5 \sqrt{2}}{2}$$ (Diagonal in a square is $$\sqrt{2}$$ times the side \& a cube has all square faces).

Finally the minimum possible volume of the cylinder can be $$\pi r^2 h=\pi \times (\frac{5 \sqrt{2}}{2})^2 \times 5=\frac{250 \pi}{4}=62.5 \pi$$ which is same as column B quantity $$\frac{125 \pi}{2}=62.5 \pi$$

As the volume of the cylinder is greater than or equal to $$62.5 \pi$$ i.e. column A quantity is greater than or equal to $62.5 \pi$, it cannot be uniquely compared with column B quantity i.e. $$62.5 \pi$$.

Hence the answer is (D).