For a moving coil galvanometer, the deflection in the coil is 0 ....

Given: $\theta = 0 . 05rad, I = 10mA, K = 4 \times {10}^{- 5}Nm{rad}^{- 1}, B = 0 . 01T, N = 200$

The torsional constant of a wire used in a moving coil galvanometer is given by

$K = \dfrac{NIAB}{\theta}$, where, $A$ is area of each turn

So, $A = \dfrac{K\theta}{NIB} = \dfrac{4 \times {10}^{- 5} \times 0 . 05}{200 \times 10 \times {10}^{- 3} \times 0 . 01}$

Thus, $A = {10}^{- 4}{m}^{2} = 1{cm}^{2}$