smc

Let the side lengths of the rectangular box be $x,y$ and $z$. From the information we get $$4(x+y+z)=48\Rightarrow x+y+z=12$$ $$2(xy+yz+xz)=94$$ The sum of all the lengths of the box's interior diagonals is $$4\sqrt{x^2+y^2+z^2}$$ Squaring the first expression, we get: $$\begin{array}{rl}144=(x+y+z{)}^2& =x^2+y^2+z^2+2(xy+yz+xz)\\ & =x^2+y^2+z^2+94.\end{array}$$ Hence $$x^2+y^2+z^2=50$$ $$4\sqrt{x^2+y^2+z^2}=\boxed{20\sqrt{2}}$$