Answer
$u = 0.028 \mathrm{~J/m^3}$
Work Step by Step
The electric field $E$ equals the potential $V$ between the two plates divided by the distance $d$ between the two plates
$$E = \dfrac{V}{d} = \dfrac{400 \,\text{V}}{0.005 \,\text{m}} = 8 \times 10^{4} \,\text{V/m} $$
Energy is the amount of work stored in the ca[acitor and the energy density $u$ is related to the electric field by
$$u = \dfrac{1}{2} \epsilon_o E^2$$
$\epsilon_o$ is the permittivity of the free space and equals $8.85 \times 10^{-12} \mathrm{~C^2/N\cdot m^2}$. Substitute to get $u$
\begin{align*}
u &= \dfrac{1}{2} \epsilon_o E^2\\
&= \dfrac{1}{2} (8.85 \times 10^{-12} \mathrm{~C^2/N\cdot m^2}) (8 \times 10^{4} \,\text{V/m})^2 \\
& =\boxed{ 0.028 \mathrm{~J/m^3}}
\end{align*}