Answer
$i_1 = 4.55~A$
Work Step by Step
A long time after the switch is closed, the inductor acts like an ordinary wire.
We can find the equivalent resistance of $R_2$ and $R_3$:
$\frac{1}{R_{23}} = \frac{1}{R_2}+\frac{1}{R_3}$
$\frac{1}{R_{23}} = \frac{1}{20.0~\Omega}+\frac{1}{30.0~\Omega}$
$\frac{1}{R_{23}} = \frac{3}{60.0~\Omega}+\frac{2}{60.0~\Omega}$
$R_{23} = 12.0~\Omega$
We can find the equivalent resistance of $R_1$ and $R_{23}$ which are in series:
$R_{eq} = R_1+R_{23} = 10.0~\Omega+12.0~\Omega = 22.0~\Omega$
We can find the total current $i$ in the circuit:
$i = \frac{100~V}{22.0~\Omega} = 4.55~A$
Since all the current goes through $R_1$, then $i_1 = 4.55~A$
