A rope is wound around a hollow cylinder of mass 3 kg and radius 40...

The torque about centre is $τ = I\alpha$, here $I$ is the moment of inertia and $\alpha$ is angular acceleration.

The formula of moment of inertia is $M{R}^{2}$.

$F \times R = M{R}^{2}\alpha$

$30 \times 0.4 = 3 \times {\left(0.4\right)}^{2}\alpha$

$12 = 3 \times 0.16\alpha$

$400 = 16\alpha$

Thus, angular acceleration is $\alpha = 25rad / {s}^{2}$