Answer
$u=34.2\frac{J}{m^3}$
Work Step by Step
We know that;
$B=\frac{\mu_{\circ}Ni}{l}$
We plug in the known values to obtain:
$B=\frac{4\times3.1416\times 10^{-7}(950)(6.60)}{0.85}=0.0092696$
We know that the energy density is given as:
$u=\frac{B^2}{2\mu_{\circ}}$
We substitute the value of $B$ found above and the value of $\mu_{\circ}$ and solve:
$u=\frac{(0.0092696)^2}{2(4\times3.1416\times 10^{-7})}=34.189=34.2\frac{J}{m^3}$
