jmc

Let $D$ be the vertex opposite $B$ on the bottom face, and let $C$ be one of the other two vertices on the bottom face. Because $BCD$ is a right triangle, we can use the Pythagorean theorem to get $BD=\sqrt{10^2+20^2}$. Then applying the Pythagorean theorem to right triangle $BDA$, we find $$\begin{array}{rl}AB& =\sqrt{10^2+{\left(\sqrt{10^2+20^2}\right)}^2}\\ & =\sqrt{10^2+10^2+20^2}\\ & =\sqrt{600}=\boxed{10\sqrt{6}}\text{ inches}.\end{array}$$