Initial speed of the automobile, $v_0 = 54 \text{ km h}^{-1} = 54 \times \frac{5}{18} \text{ m s}^{-1} = 15 \text{ m s}^{-1}$. Assuming pure rolling, the initial angular velocity of the wheels is $\omega_0 = \frac{v_0}{R} = \frac{15}{0.45} = \frac{1500}{45} = \frac{100}{3} \text{ rad s}^{-1}$. The vehicle is brought to rest, so the final angular velocity is $\omega = 0$. Time taken to come to rest, $t = 15 \text{ s}$. Using the kinematic equation for rotational motion: $\omega = \omega_0 + \alpha t$ $0 = \frac{100}{3} + \alpha(15)$ $\implies \alpha = -\frac{100}{45} = -\frac{20}{9} \text{ rad s}^{-2}$ The magnitude of angular acceleration is $|\alpha| = \frac{20}{9} \text{ rad s}^{-2}$. The magnitude of average torque is given by $\tau = I|\alpha|$. Given the moment of inertia $I = 3 \text{ kg m}^2$, $\tau = 3 \times \frac{20}{9} = \frac{20}{3} = 6.66 \text{ kg m}^2 \text{ s}^{-2}$.