In the first case, the emf of the battery is

$\mathrm{E}={\text{Kl}}_1$, where $\mathrm{K}$ is the potential gradient and ${\mathrm{l}}_1$ is the first balancing length.

In the second case the potential difference across the battery is

$\frac{\text{ER}}{\text{r}+\text{R}}={\text{Kl}}_2$, where $\mathrm{R}$ is the shunt resistance used.

Dividing the above two equations, we get

$\frac{r}{R}+1=\frac{l_1}{l_2}$

$\text{⇒r}=2\left[\frac{240-120}{120}\right]=2 \mathrm{\Omega }$