Answer
$i_1 = 0.333~A$
Work Step by Step
We can find the equivalent resistance of the three $R_2$ resistors:
$\frac{1}{R} = \frac{1}{R_2}+\frac{1}{R_2}+\frac{1}{R_2}$
$\frac{1}{R} = \frac{3}{R_2}$
$R = \frac{R_2}{3}$
$R = \frac{18.0~\Omega}{3}$
$R = 6.00~\Omega$
We can find the equivalent resistance of all four resistors:
$R_{eq} = 6.00~\Omega+R_1$
$R_{eq} = 6.00~\Omega+6.00~\Omega$
$R_{eq} = 12.0~\Omega$
We can find the current in the circuit:
$i = \frac{12.0~V}{12.0~\Omega} = 1.0~A$
Since the three $R_2$ resistors are in parallel, the current through each of them is $\frac{1.0~A}{3} = 0.333~A$
Therefore:
$i_1 = 0.333~A$
