In the given circuit diagram, the two $15\mathit{\Omega }$ resistor are in series, so their total resistance
${\mathrm{R}}_{\mathrm{}\mathit{s}}=15+15=30\mathit{\Omega }$
Now, ${\mathit{R}}_{\mathit{s}}$ and the given $30\mathit{\Omega }$ resistors are in parallel, so the net resistance
$\frac{1}{{\mathrm{R}}_{\mathit{p}}}=\frac{1}{30}+\frac{1}{30}$
$\frac{1}{{\mathrm{R}}_{\mathit{p}}}=\frac{1+1}{30}=\frac{2}{30}$
$\therefore {\mathrm{R}}_{\mathit{p}}=\frac{30}{2}=15\mathit{\Omega }$
So, ${\mathrm{R}}_{\mathit{e}\mathit{q}}=15\mathit{\Omega }$
We know, $\mathrm{I}=\frac{\mathit{\epsilon }}{{\mathrm{R}}_{\mathit{e}\mathit{q}}+\mathit{r}}$
$\therefore \mathrm{I}=\frac{3.4}{15+2}=\frac{3.4}{17}=0.2\mathrm{A}$