Answer
$t = 1.0~ns$
Work Step by Step
Note that the final value of the current is $\frac{\mathscr{E}}{R}$
We can find $t$:
$i = \frac{\mathscr{E}}{R}~(1-e^{-tR/L})$
$\frac{i~R}{\mathscr{E}} =(1-e^{-tR/L})$
$e^{-tR/L} = 1-\frac{i~R}{\mathscr{E}}$
$e^{tR/L} = \frac{1}{1-\frac{i~R}{\mathscr{E}}}$
$\frac{tR}{L} = ln(\frac{1}{1-\frac{i~R}{\mathscr{E}}})$
$t = \frac{L}{R}~{ln(\frac{1}{1-\frac{i~R}{\mathscr{E}}})}$
$t = \frac{8.0\times 10^{-6}~H}{4000~\Omega}~{ln[\frac{1}{1-\frac{(2.0\times 10^{-3}~A)~(4000~\Omega)}{20~V}}}]$
$t = \frac{8.0\times 10^{-6}~H}{4000~\Omega}~{ln(\frac{1}{1-0.4}})$
$t = 1.0~ns$
