Since the four objects are kept gently, no external torque acts on the system. Therefore, the angular momentum of the system remains conserved. Initial angular momentum of the ring, $L_i = I_i\omega_i = (Mr^2)\omega$ When four objects each of mass $m$ are placed at the opposite ends of two perpendicular diameters (i.e., at a distance $r$ from the axis), the final moment of inertia of the system becomes $I_f = Mr^2 + 4mr^2 = (M+4m)r^2$. Let the final angular velocity be $\omega_f$. By conservation of angular momentum: $L_i = L_f$ $(Mr^2)\omega = (M+4m)r^2\omega_f$ $\omega_f = \frac{M\omega}{M+4m}$