If we flow a current i, then
B $=\frac{\mu_{0}Ni}{2\pi R}$
$\phi_{self}=\left(B\times A\right)\times N=\left(\frac{\mu_{0}Ni}{2\pi R}\right)A\times N\quad\Rightarrow \quad\phi_{sell}=\left(\frac{\mu_{0}N^{2}A}{2\pi R}\right)i$ and $\phi_{self} =$ Li
$L=\frac{\mu_{0}N^{2}A}{2\pi R}=\frac{\mu_{0}}{4\pi} \frac{2N^{2}A}{R}\quad\quad\quad\quad\quad\quad\quad\quad\Rightarrow \quad L = \left(10^{-7}\right)\times\frac{2\times\left(500\right)^{2}\times10\times\left(10^{-2}\right)^{2}}{0.4}=125\mu H$
