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

The electric field at a distance $r$ from the center of the charged metal sphere ($r > R$) is given by $E = \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r^2}$.

The energy density of the electric field is $u = \dfrac{1}{2}\epsilon_0 E^2 = \dfrac{Q^2}{32\pi^2\epsilon_0 r^4}$.

The energy stored in a spherical shell of radius $r$ and thickness $dr$ is $dU = u(4\pi r^2 dr) = \dfrac{Q^2}{8\pi\epsilon_0 r^2}dr$.

Since the electric field inside the metal sphere ($r < R$) is zero, the energy $U_1$ stored inside the concentric sphere of radius $2R$ corresponds to the region $R < r < 2R$:

$U_1 = \int_R^{2R} \dfrac{Q^2}{8\pi\epsilon_0 r^2}dr = \dfrac{Q^2}{8\pi\epsilon_0}\left[-\dfrac{1}{r}\right]_R^{2R} = \dfrac{Q^2}{8\pi\epsilon_0}\left(\dfrac{1}{R} - \dfrac{1}{2R}\right) = \dfrac{Q^2}{16\pi\epsilon_0 R}$

The energy $U_2$ stored outside the sphere of radius $2R$ corresponds to the region $2R < r < \infty$:

$U_2 = \int_{2R}^{\infty} \dfrac{Q^2}{8\pi\epsilon_0 r^2}dr = \dfrac{Q^2}{8\pi\epsilon_0}\left[-\dfrac{1}{r}\right]_{2R}^{\infty} = \dfrac{Q^2}{8\pi\epsilon_0}\left(0 - \left(-\dfrac{1}{2R}\right)\right) = \dfrac{Q^2}{16\pi\epsilon_0 R}$

Therefore, the ratio $\dfrac{U_2}{U_1} = 1$.