Answer
$i_1 = 7.50~A$
Work Step by Step
We can consider the clockwise path around the circuit that includes $\mathscr{E}_4, \mathscr{E}_1, \mathscr{E}_3,$ and the two resistances at the bottom of the diagram:
$\Delta V = 5.00+20.0+5.00-i_1~R-i_1~R = 0$
$2~i_1~R = 30.0~V$
$i_1 = \frac{30.0~V}{2R}$
$i_1 = \frac{30.0~V}{(2)(2.00~\Omega)}$
$i_1 = 7.50~A$
