Answer
$a=\frac{2}{3}g$
Work Step by Step
We can show that the required linear acceleration is $\frac{2}{3}g$ as follow:
$\tau=T\times r$
$\implies T=\frac{\tau}{r}$
$\implies T=\frac{I\alpha}{r}$
But $I=\frac{1}{2}mr^2$ and $\alpha=\frac{a}{r}$
$\implies T=\frac{(\frac{1}{2}mr^2)(\frac{a}{r})}{r}$
$T=\frac{1}{2}ma$
Now the net balancing force acting downward is given as
$ma=mg-T$
$\implies ma=mg-\frac{1}{2}ma$
$\implies \frac{3}{2}ma=mg$
$a=\frac{2}{3}g$
