Cells are in series as positive terminal of one cell connected with the negative terminal of other cell.
$\begin{aligned} & \mathrm{Eq}=\mathrm{E}_1+\mathrm{E}_2 \\ & =12+6=18 \mathrm{~V} \\ & \mathrm{req}=\mathrm{r}_1+\mathrm{r}_2 \\ & =6+3 \\ & =9 \Omega\end{aligned}$
Now, $I=\frac{E}{R+r}=\frac{18}{9+4}=\frac{18}{13}=1.3 ; 1 \mathrm{~A}$
