Let $T$ be the tension in the string, $a$ is the acceleration of the mass and $\alpha$ is the angular acceleration.

Weight $mg$ is acting in a downward direction.

Now from the above figure,

$mg-T=ma ...\left(1\right)$

As torque $\tau =TR=I\alpha$

The moment of inertia for the hollow cylinder is $I=m{R}^2$

$TR=m{R}^2\times \frac{a}{R}$ (Since $\alpha =\frac{a}{R}$)

 $T=ma ...\left(2\right)$

Applying $\left(1\right)+\left(2\right)$, we get 

$\Rightarrow mg=2ma$

$a=\frac{g}{2}$