Answer
$u_B = 1.0~J/m^3$
Work Step by Step
We can write an expression for the magnetic field due to a current in a straight wire:
$B = \frac{\mu_0~i}{2\pi~r}$
We can find the energy density of the magnetic field:
$u_B = \frac{B^2}{2~\mu_0}$
$u_B = \frac{\mu_0^2~i^2}{8~\pi^2~r^2~\mu_0}$
$u_B = \frac{\mu_0~i^2}{8~\pi^2~r^2}$
$u_B = \frac{(4\pi\times 10^{-7}~H/m)~(10~A)^2}{(8~\pi^2)~(0.00125~m)^2}$
$u_B = 1.0~J/m^3$
