Answer
The direction of the current $i_2$ is to the left.
Work Step by Step
We can find the equivalent resistance of the two resistances in parallel, and the resistance marked with $i_2$ near the bottom of the diagram:
$R_{eq} = \frac{R}{2}+R = 3.00~\Omega$
We can consider the clockwise path around the circuit that includes $\mathscr{E}_4, \mathscr{E}_1, \mathscr{E}_3,$ the two resistances in parallel, and the resistance marked with $i_2$ near the bottom of the diagram:
$\Delta V = 5.00+20.0+5.00-i_2~R_{eq} = 0$
$i_2~R_{eq} = 30.0~V$
$i_2 = \frac{30.0~V}{R_{eq}}$
$i_2 = \frac{30.0~V}{3.00~\Omega}$
$i_2 = 10.0~A$
Since this path is clockwise, and the current we found has a positive value, the direction of the current $i_2$ is to the left.
