Magnetic field due to a solenoid is given by, $B={\mu }_{0}ni$.

Therefore, flux through the coil will be, $\varphi =BA={\mu }_{0}niN\pi r^2$.

According to Faraday's law of electromagnetic induction $\left|\epsilon \right|=\frac{∆\varphi }{∆t}$. Therefore,

$$\begin{gathered} i=\frac{∆\varphi }{R∆t}=\frac{{\mu }_{0}nN\pi r^2}{R}\times \frac{∆i}{∆t}=\frac{4\pi \times {10}^{-7}\times 200\times 100\times \pi \times {\left(0.03\right)}^2}{4}\times \frac{2}{0.04} \\ \Rightarrow i=9{\pi }^2 \mathrm{mA} \end{gathered}$$