Answer
$\phi=0.10\mu T.m^2$
Work Step by Step
We know that;
$L=N\frac{\phi}{I}$
This can be rearranged as:
$\phi=\frac{IL}{N}$
We plug in the known values to obtain:
$\phi=\frac{0.004\times 3.75\times 10^{-3}}{150}=0.10\times 10^{-6}T.m^2=0.10\mu T.m^2$
