A metal sphere with a radius R is charged to a potential V ....

The potential of the metal sphere is $V = \dfrac{Q}{4\pi\varepsilon_0 R}$.

The charge on the sphere is $Q = 4\pi\varepsilon_0 R V$.

The electric field at a distance $r$ from the center ($r > R$) is $E = \dfrac{Q}{4\pi\varepsilon_0 r^2}$. Since the electric field inside the metal sphere ($r < R$) is zero, the energy is stored only in the region $R < r < 2R$.

The energy density of the electric field is $u = \dfrac{1}{2}\varepsilon_0 E^2$.

The electrostatic energy stored in the region between $r = R$ and $r = 2R$ is given by integrating the energy density over the volume:

$U = \int_R^{2R} u \, dV = \int_R^{2R} \dfrac{1}{2}\varepsilon_0 \left( \dfrac{Q}{4\pi\varepsilon_0 r^2} \right)^2 4\pi r^2 \, dr$

$U = \dfrac{Q^2}{8\pi\varepsilon_0} \int_R^{2R} \dfrac{1}{r^2} \, dr = \dfrac{Q^2}{8\pi\varepsilon_0} \left[ -\dfrac{1}{r} \right]_R^{2R}$

$U = \dfrac{Q^2}{8\pi\varepsilon_0} \left( \dfrac{1}{R} - \dfrac{1}{2R} \right) = \dfrac{Q^2}{16\pi\varepsilon_0 R}$

Substituting $Q = 4\pi\varepsilon_0 R V$ into the expression for $U$:

$U = \dfrac{(4\pi\varepsilon_0 R V)^2}{16\pi\varepsilon_0 R} = \dfrac{16\pi^2\varepsilon_0^2 R^2 V^2}{16\pi\varepsilon_0 R} = \pi\varepsilon_0 R V^2$