The galvanometer deflection, when key $$K_1$$ is closed but $$K_2$$ is open, equals $$\theta_0$$ (see figure). On closing $$K_2$$ also and adjusting $$R_2$$ to $$5\Omega$$, the deflection in galvanometer becomes $$\frac{\theta_0}{5}$$. The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery]:

Deflection is directly proportional to current: $$I_g \propto \theta$$

Case 1 ($$K_1$$ closed, $$K_2$$ open): $$I_{g1} = \frac{V}{R_1 + G} = C\theta_0$$

Case 2 ($$K_1$$ closed, $$K_2$$ closed): $$R_{\text{eq}} = R_1 + \frac{G R_2}{G + R_2}$$

$$I_{\text{total}} = \frac{V}{R_1 + \frac{G R_2}{G + R_2}} = \frac{V(G + R_2)}{R_1(G + R_2) + G R_2}$$

Current dividing into galvanometer branch:

$$I_{g2} = I_{\text{total}} \times \frac{R_2}{G + R_2} = \frac{V R_2}{R_1(G + R_2) + G R_2} = C\frac{\theta_0}{5}$$

Dividing Case 1 by Case 2:

$$\frac{I_{g1}}{I_{g2}} = \frac{\frac{V}{R_1 + G}}{\frac{V R_2}{R_1 G + R_1 R_2 + G R_2}} = 5$$

$$\frac{R_1 G + R_1 R_2 + G R_2}{R_2(R_1 + G)} = 5 \implies \frac{R_1 G}{R_2(R_1 + G)} + 1 = 5$$

$$\frac{R_1 G}{R_2(R_1 + G)} = 4$$

$$\frac{220 \cdot G}{5(220 + G)} = 4 \implies 44G = 4(220 + G)$$

$$11G = 220 + G \implies 10G = 220 \implies G = 22\ \Omega$$