A moving-coil galvanometer has 150 turns…

Step 1. The factor. $NAB = 150 \times 6.0 \times 10^{-4}\ \text{m}^2 \times 0.12\ \text{T} = 1.08 \times 10^{-2}\ \text{N m/A}$.

Step 2. The current sensitivity. $\dfrac{\phi}{I} = \dfrac{NAB}{k} = \dfrac{1.08 \times 10^{-2}}{1.8 \times 10^{-6}\ \text{N m/rad}} = 6000\ \text{rad/A} = 6.0\ \text{rad/mA}$.

Step 3. The full-scale current. $\phi = 45^\circ = 0.785$ rad. So $I_g = \dfrac{\phi}{6000\ \text{rad/A}} = 1.31 \times 10^{-4}\ \text{A} = 131\ \mu\text{A}$.

Step 4. The voltage. $V = I_gG = 1.31 \times 10^{-4}\ \text{A} \times 85\ \Omega = 0.0111\ \text{V} = 11.1\ \text{mV}$.