Answer
$\approx 0.18\;\rm J/m^3$
Work Step by Step
We know that the energy density in the electric field is given by
$$u =\dfrac{\epsilon_0E^2}{2}\tag 1$$
where $E$ is the electric field.
We also know that the electric field at the surface of a sphere is given by
$$E=\dfrac{k_eQ}{R^2}$$
where the potential at the surface of the sphere is given by $V=\dfrac{k_eQ}{R^2}$, so
$$E=\dfrac{k_eQ}{R }\dfrac{1}{R}=\dfrac{V}{R}$$
Plug into (1).
$$u =\dfrac{\epsilon_0V^2}{2R^2} $$
Plug the known;
$$u =\dfrac{(8.85\times 10^{-12})(1000)^2}{2(0.5\times 10^{-2})^2} $$
$$u=\color{red}{\bf 0.177}\;\rm J/m^3$$