1. Given data

• Area of loop: $A = 3.5 \times 10^{-3}\ \text{m}^2$
• Resistance: $R = 10\ \Omega$
• Magnetic field: $B(t) = (0.4)\sin(50\pi t)\ \text{T}$
• Time interval: $t: 0 \to 10\ \text{ms} = 0.01\ \text{s}$

The loop is perpendicular to the magnetic field, so magnetic flux is $\Phi(t) = AB(t)$

2. Induced emf and current

By Faraday's law, $\mathcal{E} = -\frac{d\Phi}{dt}$

Current in the loop is $I = \frac{\mathcal{E}}{R}$

The net charge flown is $q = \int_0^{0.01} I\,dt = \frac{1}{R}\int_0^{0.01} \mathcal{E}\,dt$

Using Faraday's law, $q = -\frac{1}{R}\int_0^{0.01} \frac{d\Phi}{dt}\,dt = -\frac{1}{R}[\Phi(0.01)-\Phi(0)]$

So in magnitude, $|q| = \frac{|\Phi(0.01)-\Phi(0)|}{R}$

3. Compute flux at the two instants

At $t=0$, $B(0)=0.4\sin 0=0$ Thus, $\Phi(0)=A\cdot 0=0$

At $t=0.01\ \text{s}$, $B(0.01)=0.4\sin(50\pi \times 0.01)=0.4\sin\left(\frac{\pi}{2}\right)=0.4\ \text{T}$

Hence, $\Phi(0.01)=A B = (3.5\times 10^{-3})(0.4)=1.4\times 10^{-3}\ \text{Wb}$

4. Net charge

Therefore, $|q|=\frac{1.4\times 10^{-3}}{10}=1.4\times 10^{-4}\ \text{C}$

Convert to mC: $1.4\times 10^{-4}\ \text{C}=0.14\times 10^{-3}\ \text{C}=0.14\ \text{mC}$

5. Match with options

$\boxed{q \approx 0.14\ \text{mC}}$

So the correct option is A.