Step 1: Write the solenoid field.
The field inside the solenoid is $B = \mu_0 n I$, where the turn density is \[ n = 200\ \text{turns/cm} = 2 \times 10^{4}\ \text{turns/m} \]
Step 2: Identify the flux-carrying area.
The flux through the coil is limited to the solenoid cross-section of radius $2\ \text{cm}$, so \[ A = \pi (2 \times 10^{-2})^2 = 4\pi \times 10^{-4}\ \text{m}^2 \]
Step 3: Compute the rate of change of the field.
As the current falls from $2\ \text{A}$ to $0$ in $0.04\ \text{s}$, \[ \frac{dB}{dt} = \mu_0 n \frac{dI}{dt} = (4\pi \times 10^{-7})(2 \times 10^{4})\frac{2}{0.04} \]
Step 4: Find the induced emf.
With $N = 100$ turns, the emf is \[ \varepsilon = N A \frac{dB}{dt} \] Combining all factors gives $\varepsilon = 16\pi^2 \times 10^{-3}\ \text{V}$.
Step 5: Evaluate the emf factors.
Here $\mu_0 n \dfrac{dI}{dt} = 4\pi \times 10^{-7} \times 2 \times 10^{4} \times 50 = 0.4\pi$, and multiplying by $N A = 100 \times 4\pi \times 10^{-4}$ confirms $\varepsilon = 16\pi^2 \times 10^{-3}\ \text{V}$.
Step 6: Get the induced current.
Dividing by the coil resistance $R = 4\ \Omega$, \[ i = \frac{\varepsilon}{R} = \frac{16\pi^2 \times 10^{-3}}{4} = 4\pi^2 \times 10^{-3}\ \text{A} = 4\pi^2\ \text{mA} \] so \[ \boxed{4\pi^2\ \text{mA}} \]