Answer
$\frac{\mathscr{E}_L}{\mathscr{E}} = 1$
Work Step by Step
We can find an expression for $\frac{di}{dt}$:
$i = \frac{\mathscr{E}}{R}~(1-e^{-tR/L})$
$\frac{di}{dt} = \frac{\mathscr{E}}{L}~e^{-tR/L}$
We can find an expression for $\mathscr{E}_L$:
$\mathscr{E}_L = -L~\frac{di}{dt}$
$\mathscr{E}_L = -(L)~(\frac{\mathscr{E}}{L}~e^{-tR/L})$
$\mathscr{E}_L = -\mathscr{E}~e^{-tR/L}$
Just after $t = 0$, then $\mathscr{E}_L = -\mathscr{E}$
The magnitude of $\mathscr{E}_L$ is $\mathscr{E}$
Therefore:
$\frac{\mathscr{E}_L}{\mathscr{E}} = 1$
