Classical inequalities is still relevant here!

Let $x$ , $y$ , and $z$ be the sides of the cuboid, and the face with sides $x$ and $y$ be the unpainted face. Then the volume is $V = xyz$ , the painted area is $P = xy + 2xz + 2yz = 252$ , and the area of the unpainted face is $U = xy$ .

Using the AM-GM inequality on $xy\sqrt[3]{xyz}$ , $2xz\sqrt[3]{xyz}$ , and $2yz\sqrt[3]{xyz}$ , we have $\frac{xy\sqrt[3]{xyz} + 2yz\sqrt[3]{xyz} + 2xz\sqrt[3]{xyz}}{3} \geq \sqrt[3]{4x^3y^3z^3}$ or $\frac{xy\sqrt[3]{xyz}}{3\sqrt[3]{4}} + \frac{2yz\sqrt[3]{xyz}}{3\sqrt[3]{4}} + \frac{2xz\sqrt[3]{xyz}}{3\sqrt[3]{4}} \geq V$ . The maximum value of $V$ occurs at the equality $\frac{xy\sqrt[3]{xyz}}{3\sqrt[3]{4}} = \frac{2yz\sqrt[3]{xyz}}{3\sqrt[3]{4}} = \frac{2xz\sqrt[3]{xyz}}{3\sqrt[3]{4}}$ or when $xy = 2yz = 2xz$ .

Therefore, since $xy + 2xz + 2yz = 252$ and $xy = 2yz = 2xz$ , we have $3xy = 252$ , and since $U = xy$ , we have $3U = 252$ , which means $U = \boxed{84}$ .