Explanation

(b) Here, Moment of inertia, $I = 3 \times 10^{2}kg\ m^{2}$
Torque, $\tau = 6.9 \times 10^{2}Nm$
Initial angular speed, $\omega_{0} = 4.6\ rad\ s^{- 1}$
Final angular speed, $\omega = 0\ rad\ s^{- 1}$
As $\omega = \omega_{0} + \alpha t$
$\alpha = \frac{\omega - \omega_{0}}{t} = \frac{0 - 4.6}{t} = - \frac{4.6}{t}rad\ s^{- 2}$
Negative sign is for deceleration Torque, $\tau = I\alpha$
$6.9 \times 10^{2} = 3 \times 10^{2} \times \frac{4.6}{t}$
$t = \frac{3 \times 10^{2} \times 4.6}{6.9 \times 10^{2}} = 2s$