Answer
$0.010m^2$
Work Step by Step
As $U=\frac{1}{2}LI^2$
This can be rearranged as:
$L=\frac{2U}{I^2}$
$\implies L=\frac{2(0.31)}{(12)^2}=0.0043H$
Now we can find the cross sectional area of the solenoid as follows:
$L=\mu_{\circ}n^2Al$
This can be rearranged as:
$A=\frac{L}{\mu_{\circ}n^2l}$
We plug in the known values to obtain:
$A=\frac{0.0043}{4\pi \times 10^{-7}(470)(1.5)}$
$A=0.010m^2$
