The correct answer is $\frac{G}{49}$.
The shunt is a parallel resistor that is used to divert most of the current away from the galvanometer. The current through the galvanometer is then only a small fraction of the main current. The resistance of the shunt is given by the following equation:
$$R_s = \frac{I_g}{I – I_g}R_g$$
where $I_g$ is the current through the galvanometer, $I$ is the main current, and $R_g$ is the resistance of the galvanometer.
In this case, $I_g = 0.02I$, so the resistance of the shunt is:
$$R_s = \frac{0.02I}{I – 0.02I}R_g = \frac{0.02I}{0.98I}R_g = \frac{G}{49}$$
Option A is incorrect because it is the resistance of the galvanometer, not the resistance of the shunt. Option B is incorrect because it is the reciprocal of the correct answer. Option C is incorrect because it is the resistance of the galvanometer multiplied by 49. Option D is incorrect because it is the resistance of the galvanometer.