Let the resistance of the voltmeter be RΩ. The voltmeter is in parallel with the 5Ω. So the total resistance in the circuit is
(5+R5R+2)Ω
Applying Ohm's law in the whole circuit, V=IR,
I=(2+5+R5R3)A
It is given that the potential difference across 5Ω is 2V.
Thus, by Ohm's law,
(2+5+R5R3)×(R+55R)=2⇒10+7R15+3R×5+R5R=2⇒15R=20+14R⇒R=20Ω
