A large conducting plane possesses a surface charge density of 1.0...

The electric field just outside a large conducting plane is given by:

$E = \dfrac{\sigma}{\epsilon_0}$

where $\sigma$ is the surface charge density.

The electrostatic energy density (energy stored per unit volume) in the electric field is:

$u = \dfrac{1}{2} \epsilon_0 E^2 = \dfrac{1}{2} \epsilon_0 \left(\dfrac{\sigma}{\epsilon_0}\right)^2 = \dfrac{\sigma^2}{2\epsilon_0}$

The volume of the cubical region is:

$V = a^3 = (1.0 \times 10^{-2} \text{ m})^3 = 1.0 \times 10^{-6} \text{ m}^3$

The total electrostatic energy stored in this volume is:

$U = u \times V = \dfrac{\sigma^2}{2\epsilon_0} V$

Substituting the given values $\sigma = 1.0 \times 10^{-4} \text{ C m}^{-2}$ and $\epsilon_0 = 8.85 \times 10^{-12} \text{ F m}^{-1}$:

$U = \dfrac{(1.0 \times 10^{-4})^2}{2 \times 8.85 \times 10^{-12}} \times 1.0 \times 10^{-6}$

$U = \dfrac{10^{-8} \times 10^{-6}}{17.7 \times 10^{-12}} = \dfrac{10^{-14}}{17.7 \times 10^{-12}} = \dfrac{10^{-2}}{17.7}$

$U \approx 5.6 \times 10^{-4} \text{ J}$