Answer
$L = 97.9~H$
Work Step by Step
We can find the inductance of the coil:
$i = \frac{\mathscr{E}}{R}~(1-e^{-tR/L})$
$\frac{i~R}{\mathscr{E}} = 1-e^{-tR/L}$
$e^{-tR/L} = 1-\frac{i~R}{\mathscr{E}}$
$e^{tR/L} = \frac{1}{1-\frac{i~R}{\mathscr{E}}}$
$\frac{tR}{L} = ln(\frac{1}{1-\frac{i~R}{\mathscr{E}}})$
$L = \frac{tR}{ln(\frac{1}{1-\frac{i~R}{\mathscr{E}}})}$
$L = \frac{(5.00\times 10^{-3}~s)(10,000~\Omega)}{ln(\frac{1}{1-\frac{(2.00\times 10^{-3}~A)(10,000~\Omega)}{50.0~V}})}$
$L = \frac{(5.00\times 10^{-3}~s)(10,000~\Omega)}{ln(\frac{1}{1-0.400})}$
$L = \frac{(5.00\times 10^{-3}~s)(10,000~\Omega)}{0.510826}$
$L = 97.9~H$
