Answer
$i_1 = 3.33~A$
Work Step by Step
When the switch is first closed, the inductor acts like a broken wire, so all the current will flow through $R_1$ and then through $R_2$.
We can find the equivalent resistance of $R_1$ and $R_2$:
$R_{12} = R_1+R_2 = 10.0~\Omega+20.0~\Omega = 30.0~\Omega$
We can find the initial current in the circuit:
$i = \frac{\mathscr{E}}{R_{12}} = \frac{100~V}{30.0~\Omega} = 3.33~A$
Therefore:
$i_1 = 3.33~A$
