Answer
$\alpha=8.18rad/s^2$
Work Step by Step
We have $$\sum\tau=I\alpha$$ $$\alpha=\frac{\sum\tau}{I}$$
The net torque $\sum\tau=1.8N.m$, and the blades' moment of inertia $I=0.22kg.m^2$. The angular acceleration of the blades is $$\alpha=8.18rad/s^2$$
