To solve this problem, we need to determine the acceleration of the blocks considering the pulley system described. Let's go step-by-step through the solution:

1. **Understand the Mass Distribution of the Pulley:**
   • The pulley consists of a thin rim and two rods forming a cross, each with mass \(M\).
   • The total mass of the pulley is \(3M\) (since it comprises the rim and two rods).
2. **Moment of Inertia of the Pulley:**
   • The moment of inertia for the thin rim about its center is \(I_{\text{rim}} = MR^2\).
   • Each rod has a moment of inertia \(\frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2 = \frac{1}{3}ML^2\) when treated as a rod about the center.
   • Thus, the moment of inertia for both rods is \(\frac{2}{3}ML^2\).
   • The effective moment of inertia of the entire pulley system is \(I = MR^2 + \frac{2}{3}MR^2 = \frac{5}{3}MR^2\).
3. **Apply Newton's Second Law:**
   • The tension in the string causes torques on the pulley. Let tensions be \(T_1\) and \(T_2\) on blocks \(M\) and \(m\) respectively.
   • For the block of mass \(M\), applying Newton's second law gives: \(Mg - T_1 = Ma\).
   • For the block of mass \(m\), applying Newton's second law gives: \(T_2 - mg = ma\).
   • For the pulley, using torque: \((T_1 - T_2)R = I\alpha\). The angular acceleration \(\alpha = \frac{a}{R}\) (since there is no slipping).
4. **Calculation:**
   • Substitute \(I = \frac{5}{3}MR^2\) and \(\alpha = \frac{a}{R}\) into the torque equation: \((T_1 - T_2)R = \frac{5}{3}MRa\).
   • Hence, \(T_1 - T_2 = \frac{5}{3}Ma\).
   • Using the three equations for forces and torque, solve them simultaneously to find:
     • \(T_1 = M(g - a)\)
     • \(T_2 = m(g + a)\)
   • Substitute these in the torque equation and solve for \(a\):
   • After substitution and simplification, the acceleration \(a\) is found to be: \(a = \frac{(M - m)g}{\left[\frac{8}{3}M + m\right]}\).

Therefore, the correct answer is \(\frac{(M - m)g}{\left[\frac{8}{3}M + m\right]}\).