Let block falls through a height $h$, then loss of $\mathrm{PE}$ of block appears in form of KE of block and KE of rotation of pulley.
$\Rightarrow \underset{\text { (block })}{m g h}=\underset{\text { (block) }}{\frac{1}{2}} m v^2+\underset{\text { (pulley) }}{\frac{1}{2}} I \omega^2$
$\Rightarrow m g h=\frac{1}{2} m v^2+\frac{1}{2} \frac{M R^2}{2} \cdot \omega^2$
Here, $m=M=2 \mathrm{~kg}$,
So, $g h=\frac{v^2}{2}+\frac{R^2}{4} \omega^2$
As $v=R \omega, g h=\frac{3}{4} R^2 \omega^2$
Differentiating with respect to time,
$g \cdot \frac{d h}{d t}=\frac{3}{4} R^2 \cdot 2 \omega \cdot \frac{d \omega}{d t}$
$g v=\frac{3}{4} R^2 \times 2 \times \frac{v}{R} \cdot \alpha$
$\Rightarrow \quad g=\frac{6}{4} R \alpha \Rightarrow \frac{4 g}{6}=R \alpha \Rightarrow R \alpha=\frac{2 g}{3}$
But $R \alpha=a_t=$ tangential acceleration of pulley
$=$ acceleration of block
So, $a_t=R \alpha=\frac{2}{3} g=\frac{20}{3} \mathrm{~m} / \mathrm{s}^2$