jmc

Let the dimensions of $P$ be $x$, $y$, and $z$. The sum of the lengths of the edges of $P$ is $4(x+y+z)$, and the surface area of $P$ is $2xy+2yz+2xz$, so $$x+y+z=28\text{and}2xy+2yz+2xz=384.$$ Each internal diagonal of $P$ is a diameter of the sphere, so $$\begin{array}{rl}(2r{)}^2& =(x^2+y^2+z^2)\\ & =(x+y+z{)}^2-(2xy+2xz+2yz)\\ & =28^2-384\\ & =400.\end{array}$$ So $2r=20$ and $r=\boxed{10}$.

Note: There are infinitely many positive solutions of the system $x+y+z=28$, $2xy+2yz+2xz=384$, so there are infinitely many non-congruent boxes meeting the given conditions, but each can be inscribed in a sphere of radius 10.