In the arrangement shown in the figure, the two identical pulleys...

Let $a$ be the downward acceleration of block $M$. Since the string is inextensible, block $m$ will have an upward acceleration $a$.

Let $T_1$ be the tension in the string attached to $M$, $T_2$ be the tension in the horizontal segment of the string between the two pulleys, and $T_3$ be the tension in the string attached to $m$.

Writing the equations of motion for the two blocks:
For block $M$:
$Mg - T_1 = Ma$ ... (i)

For block $m$:
$T_3 - mg = ma$ ... (ii)

For the two pulleys, assuming no slipping, the angular acceleration is $\alpha = \dfrac{a}{R}$.
Writing the torque equations for the two identical pulleys:
For the first pulley (above $M$):
$(T_1 - T_2)R = I\alpha = I\left(\dfrac{a}{R}\right)$
$\Rightarrow T_1 - T_2 = \dfrac{Ia}{R^2}$ ... (iii)

For the second pulley (above $m$):
$(T_2 - T_3)R = I\alpha = I\left(\dfrac{a}{R}\right)$
$\Rightarrow T_2 - T_3 = \dfrac{Ia}{R^2}$ ... (iv)

Adding equations (iii) and (iv), we get:
$T_1 - T_3 = \dfrac{2Ia}{R^2}$ ... (v)

From equations (i) and (ii), we can substitute the values of $T_1$ and $T_3$ into equation (v):
$(Mg - Ma) - (mg + ma) = \dfrac{2Ia}{R^2}$

Rearranging the terms to solve for $a$:
$(M - m)g - (M + m)a = \dfrac{2Ia}{R^2}$

$(M - m)g = a\left(M + m + \dfrac{2I}{R^2}\right)$

$a = \dfrac{(M - m)g}{M + m + \dfrac{2I}{R^2}}$