Answer
$U_{total}=1.98\mu J$
Work Step by Step
We know that;
$U_{total}=\frac{1}{2}LI^2+\frac{1}{2}\frac{Q^2}{C}$
We plug in the known values to obtain:
$U_{total}=\frac{1}{2}(0.025)(0.00920)^2+\frac{1}{2}\frac{(3.80\times 10^{-6})^2}{7.80\times 10^{-6}}
=1.9836\times 10^{-6}=1.98\mu J$
