Answer
$10.6~J/m^3$
Work Step by Step
We can find the electric energy density as follows:
$E=\frac{1}{2}CV^2$
We plug in the known values to obtain:
$E=\frac{1}{2}(225\times 10^{-6}F)(345V)^2$
$E=13.4J$
and $A=\frac{Cd}{\epsilon_{\circ}}$
$A=\frac{(225\times 10^{-6}F)(0.223\times 10^{-3}m)}{8.85\times 10^{-12}F/m}$
$A=5.66949\times 10^3m^2$
Now the electric energy density is given as
$Electric \space energy\space density=\frac{E}{A\times t}$
We plug in the known values to obtain:
$Electric \space energy\space density=\frac{13.4J}{(5.66949m^2)(0.223\times 10^{-3}m)}=10.6~J/m^3$
