Each pulley in the figure has radius r and moment of inertia I. The acceleration of the block is 

 

$T_1-mg=ma$……..(i)

$\begin{array}{r}r\left(T_2-T_1\right)=I\alpha .........\left(ii\right)\\ Mg-T_3=Ma.........\left(iii\right)\\ r\left(T_3-T_2\right)=I\alpha .........\left(iv\right)\end{array}$

and a = R$\alpha$

From Eqs. (ii) and (iv), we get

$T_3-T_1=\frac{2I\alpha }{r}=\frac{2Ia}{R^2}$

From Eqs. (i) and (iii), we get

$\begin{array}{l}(M-m)g=(M+m)a+T_3-T_1\\ \Rightarrow (M-m)g=(M+m)a+\frac{2Ia}{r^2}\\ \Rightarrow a=\frac{(M-m)g}{\left(M+m+\frac{2I}{r^2}\right)}\end{array}$