1. Initial moment of inertia of the ring

A thin circular ring of mass $m$ and radius $R$ rotating about its central axis has

$I_i = mR^2$

2. Initial angular momentum

Since the ring is rotating with angular velocity $\omega$, its initial angular momentum is

$L_i = I_i\omega = mR^2\omega$

3. Final moment of inertia after attaching two masses

Two particles, each of mass $M$, are attached at opposite ends of a diameter. Each particle is at distance $R$ from the axis.

Moment of inertia of the two particles:

$I_{\text{particles}} = MR^2 + MR^2 = 2MR^2$

So the total final moment of inertia is

$I_f = mR^2 + 2MR^2 = (m+2M)R^2$

4. Use conservation of angular momentum

Since the masses are attached gently and no external torque acts about the axis,

$L_i = L_f$

So,

$mR^2\omega = (m+2M)R^2\omega'$

Cancel $R^2$:

$m\omega = (m+2M)\omega'$

Hence,

$\omega' = \frac{m\omega}{m+2M}$

5. Match with the options

This corresponds to

$\boxed{\omega' = \frac{\omega m}{m+2M}}$

So the correct option is D.