A coil of 150 turns and radius 2.0 cm is placed with its axis along the field of an electromagnet…

Step 1. The change of flux. $A = \pi \times (0.020\ \text{m})^2 = 1.257 \times 10^{-3}\ \text{m}^2$. So $\Delta\Phi = A\,\Delta B = 1.257 \times 10^{-3}\ \text{m}^2 \times 0.80\ \text{T} = 1.005 \times 10^{-3}\ \text{Wb}$.

Step 2. The emf. $\varepsilon = N\dfrac{\Delta\Phi}{\Delta t} = 150 \times \dfrac{1.005 \times 10^{-3}\ \text{Wb}}{0.25\ \text{s}} = 0.603\ \text{V}$.

Step 3. The current. $I = \dfrac{\varepsilon}{R} = \dfrac{0.603\ \text{V}}{12\ \Omega} = 0.0503\ \text{A}$, that is $50.3$ mA.

Step 4. The charge. $Q = I\,\Delta t = 0.0503\ \text{A} \times 0.25\ \text{s} = 0.0126\ \text{C}$, that is $12.6$ mC.