Answer
$\frac{\mathscr{E}_L}{\mathscr{E}} = \frac{1}{e^2}$
Work Step by Step
We can find an expression for $\frac{di}{dt}$:
$i = \frac{\mathscr{E}}{R}~(1-e^{-t/\tau_L})$
$\frac{di}{dt} = \frac{\mathscr{E}}{L}~e^{-t/\tau_L}$
We can find an expression for $\mathscr{E}_L$:
$\mathscr{E}_L = -L~\frac{di}{dt}$
$\mathscr{E}_L = -(L)~(\frac{\mathscr{E}}{L}~e^{-t/\tau_L})$
$\mathscr{E}_L = -\mathscr{E}~e^{-t/\tau_L}$
When $t = 2.00~\tau_L$, then $\mathscr{E}_L = -\frac{\mathscr{E}}{e^2}$
The magnitude of $\mathscr{E}_L$ is $\frac{\mathscr{E}}{e^2}$
Therefore:
$\frac{\mathscr{E}_L}{\mathscr{E}} = \frac{1}{e^2}$
