Given data

• Galvanometer resistance: $G = 50\,\Omega$
• Maximum current through galvanometer: $I_g = 5\text{ mA} = 5 \times 10^{-3}\text{ A}$
• Desired voltmeter range: $V = 100\text{ V}$

Condition for conversion into voltmeter To convert a galvanometer into a voltmeter of range $V$, a high resistance $R$ is connected in series such that at full-scale deflection, $V = I_g (G + R)$

Substitute the values $100 = 5 \times 10^{-3}(50 + R)$

Solve for $R$ $50 + R = \frac{100}{5 \times 10^{-3}}$ $50 + R = \frac{100}{0.005} = 20000$ $R = 20000 - 50 = 19950\,\Omega$

Match with options The required series resistance is $\boxed{19950\,\Omega}$ which corresponds to Option D.