The voltmeter depicted in the figure measures a potential...

Given the potential difference across the parallel combination is $V_p = 18\text{ V}$.

The potential difference across the $24\text{ }\Omega$ resistor is:
$V_{24} = 30 - 18 = 12\text{ V}$

The current in the main circuit is:
$I = \dfrac{12}{24} = 0.5\text{ A}$

Let $R_V$ be the resistance of the voltmeter. The equivalent resistance of the parallel part is:
$R_p = \dfrac{18}{0.5} = 36\text{ }\Omega$

For the parallel combination:
$\dfrac{1}{R_p} = \dfrac{1}{50} + \dfrac{1}{R_V}$

$\dfrac{1}{36} = \dfrac{1}{50} + \dfrac{1}{R_V}$

$\dfrac{1}{R_V} = \dfrac{1}{36} - \dfrac{1}{50} = \dfrac{25 - 18}{900} = \dfrac{7}{900}$

$R_V = \dfrac{900}{7} \approx 128.6\text{ }\Omega$

The closest option to this value is $130\text{ }\Omega$.

Answer: $130 \text{ }\Omega$