Given: Mass of the solid cylinder, $m = 2 \text{ kg}$ Radius, $r = 4 \text{ cm} = 0.04 \text{ m}$ Initial angular velocity, $\omega_0 = 3 \text{ rpm} = 3 \times \frac{2\pi}{60} \text{ rad/s} = \frac{\pi}{10} \text{ rad/s}$ Final angular velocity, $\omega = 0$ Angular displacement, $\theta = 2\pi \text{ revolutions} = 2\pi \times 2\pi \text{ rad} = 4\pi^2 \text{ rad}$

Using the equation of rotational kinematics: $\omega^2 = \omega_0^2 + 2\alpha\theta$ $0 = \left(\frac{\pi}{10}\right)^2 + 2\alpha(4\pi^2)$ $2\alpha(4\pi^2) = -\frac{\pi^2}{100}$ $\alpha = -\frac{1}{800} \text{ rad/s}^2$

The moment of inertia of the solid cylinder about its axis: $I = \frac{1}{2}mr^2 = \frac{1}{2} \times 2 \times (0.04)^2 = 0.0016 \text{ kg m}^2 = 16 \times 10^{-4} \text{ kg m}^2$

The magnitude of the required torque is: $\tau = I|\alpha| = (16 \times 10^{-4} \text{ kg m}^2) \times \left(\frac{1}{800} \text{ rad/s}^2\right) = 2 \times 10^{-6} \text{ N-m}$