To convert a galvanometer into a voltmeter to measure higher voltages, you need to add a series resistance to it. Let's figure out the required resistance value.

First, let's find the maximum voltage that can be directly measured by the galvanometer without any additional resistance. We know the maximum current, $I$, that the galvanometer can safely measure is $5 \text{ mA}$, and the resistance of the galvanometer, $R_g$, is $50 \Omega$.

Using Ohm's law $ V = I \times R $, the maximum voltage $V_g$ the galvanometer can measure is:

$ V_g = I \times R_g $

$ V_g = 5 \times 10^{-3} \text{ A} \times 50 \Omega $

$ V_g = 0.25 \text{ V} $

Next, to measure up to $100 \text{ V}$, we need to add a series resistor $R_s$ so that the total voltage drop when the maximum current is flowing is $100 \text{ V}$. The voltage drop across the additional resistor $R_s$ when the maximum current flows would be the total voltage minus the voltage across the galvanometer:

$ V_s = V_{\text{total}} - V_g $

$ V_s = 100 \text{ V} - 0.25 \text{ V} $

$ V_s = 99.75 \text{ V} $

Applying Ohm's law to the series resistor to find its resistance value:

$ R_s = \frac{V_s}{I} $

$ R_s = \frac{99.75 \text{ V}}{5 \times 10^{-3} \text{ A}} $

$ R_s = 19950 \Omega $

Therefore, the resistance of the series resistor required to convert the galvanometer into a voltmeter that can measure up to $100 \text{ V}$ is $19950 \Omega$.

The correct option is D: $19950 \Omega$.