jmc

Let $a,$ $b,$ and $c$ be the dimensions of the box, so $abc=216.$ Then the surface area is $$2(ab+ac+bc).$$By AM-GM, $$ab+ac+bc\ge 3\sqrt[3]{(ab)(ac)(bc)}=3\sqrt[3]{a^2{b}^2{c}^2}=108,$$so $2(ab+ac+bc)\ge 216.$

Equality occurs when $a=b=c=6,$ so the smallest possible surface area is $\boxed{216}.$