Solution

In electrostatics, the energy stored in the electric field produced by a charge distribution is equal to the potential energy of the distribution. If a ball of radius $r$ has uniform charge density $\rho$, then the potential at its surface is

$$V=\frac{k}{r}\cdot\rho\cdot\frac{4}{3}\pi r^3=\frac{\rho r^2}{3\epsilon_0}.$$

If a thin spherical shell of charge $\rho\cdot4\pi r^2\,dr$ is added to its surface, then the potential energy of the system increases by

$$dU=V\cdot\rho\cdot4\pi r^2\,dr=\frac{4\pi\rho^2r^4}{3\epsilon_0}.$$

(The potential energy in the thin spherical shell is negligible compared to $dU$.) Thus,

$$U=\frac{4\pi\rho^2r^5}{15\epsilon_0}=\frac{3k}{5r}\left(\rho\cdot\frac{4}{3}\pi r^3\right)^2.$$

Using $R$ for $r$ and $Q$ for $\rho\cdot4\pi R^3/3$, we have $U=3kQ^2/(5r)$.