Answer
$u = 9.15\times 10^{-18}~J/m^3$
Work Step by Step
We can find the magnitude of the electric field at $r = 1.00~mm$:
$E = \frac{1}{4\pi~\epsilon_0}~\frac{\vert q \vert}{r^2}$
$E = \frac{1}{(4\pi)~(8.854\times 10^{-12}~C/V~m)}~\frac{1.6\times 10^{-19}~C}{(1.00\times 10^{-3}~m)^2}$
$E = 1.438\times 10^{-3}~V/m$
We can find the energy density:
$u = \frac{1}{2}~\epsilon_0~E^2$
$u = \frac{1}{2}~(8.854\times 10^{-12}~C/V~m)(1.438\times 10^{-3}~V/m)^2$
$u = 9.15\times 10^{-18}~J/m^3$