Answer
$\alpha=182.5rad/s^2$ and $T=5N$
Work Step by Step
Let's examine the block. Its weight and tension in the cord are the forces acting on it. Because it is falling with acceleration $a$, we have $$W-T=ma$$
The cord has a rolling motion without slipping, so $a=r_{cord}\alpha$
$$W-T=mr\alpha (1)$$ $$W-T=mr\frac{\sum\tau}{I}$$
Tension $T$ is the only force producing torque as the axis of rotation is at the center of the pulley. The lever arm is the radius of the cord $r$ $$W-T=mr\frac{Tr}{I}=\frac{mTr^2}{I}$$ $$W=T\Big(1+\frac{mr^2}{I}\Big)=T\Big(\frac{I+mr^2}{I}\Big)$$ $$T=\frac{WI}{I+mr^2}$$
We have the block's weight $W=2\times9.8=19.6N$, $I=1.1\times10^{-3}kg.m^2$, the block's mass $m=2kg$ and the cord's radius $r=0.04m$
$$T=5N$$
To find $\alpha$, we go back to (1): $$\alpha=\frac{W-T}{mr}=182.5rad/s^2$$
