b
Let $\varepsilon$ be emf and $\mathrm{r}$ be intemal resistance of the battery.
In first case, $12=\varepsilon-2 \mathrm{r}$ ..........$(i)$
In second case, $15=\varepsilon+3 \mathrm{r}$ .........$(ii)$
Substract $(ii)$ from $(i),$ we get, $\mathrm{r}=\frac{3}{5}\, \Omega$
Putting this value of $\mathrm{r}$ in equation $(\mathrm{i}),$ we get
$\varepsilon=12+\frac{2 \times 3}{5}=\frac{60+6}{5}=\frac{66}{5}=13.2 \mathrm{\,V}$