Two resistance of 100Ω and 200Ω are connected in series with a...

Voltage across $200Ω = 4 - 1 = 3V$.

Therefore, current through the battery, $\dfrac{3}{200}$

Now, equivalent resistance of ${R}_{V} \text{ and } 100Ω$,$= \dfrac{{R}_{v}100}{{R}_{v} + 100}$

For voltmeter, $\dfrac{{R}_{v}100}{{R}_{v} + 100} \times \dfrac{3}{200} = 1$

$⇒ 3{R}_{v} = 2{R}_{v} + 200$

$⇒ {R}_{v} = 200Ω$