 Solution

Spherical symmetry and the integral Gauss law give $4\pi r^2{E}_r(r)=Q/{ϵ}_0$ for $r>R$. Together with the stipulated zero interior field,The electric field jumps across the ideal charged surface; its value exactly on that surface is not needed for the volume integral. The electrostatic energy of a uniformly charged spherical shell isAs a check, the shell has constant electric potential $Q/(4\pi {ϵ}_{0}R)$, with zero reference at infinity, so $U=Q\varphi (R)/2$ gives the same result.