Given: Initial number of cells $\left(n_1\right)=10$;

Potential of each cell $=E$; Internal resistance of each cell $=r$ and final number of cells $\left(n_2\right)=3$

We know from the Ohm's law, total voltage of ten cells $=10 \times E=10 E$ and total resistance in ten cells $=10 \times r$ $=10\,r$

Therefore current in the circuit $(I)=\frac{10 E}{10 r}=\frac{E}{r}$ or potential difference across three cells $=I \times 3 r=\frac{E}{r} \times 3 r=3\,E$

(Since the voltmeter is ideal, therefore it will read $3E$ ).