The current ' i ' in the circuit shown in the figure is

In the given question, Resistance in series with $2\epsilon$ is not given, assuming it to be $R$.

The above circuit can be redrawn as,

Using equivalent battery for the dotted portion shown in the figure below,

${\epsilon }_{eq}=\frac{{\displaystyle \frac{\epsilon }{R}}+{\displaystyle \frac{2\epsilon }{R}}}{{\displaystyle \frac{1}{R}}+{\displaystyle \frac{1}{R}}}=1.5\epsilon$ and $R_{eq}=\frac{R}{2}$. Therefore,

$i=\frac{\left(3-1.5\right)\epsilon }{\left(2.5+0.5\right)R}=\frac{\epsilon }{2R}$