Step-by-step Solution:

1. Identify the Physical Principle The system consists of a ring and two point masses. The masses move radially outwards, which is due to internal forces within the system. There are no external torques acting on the system about the vertical axis of rotation. Therefore, the angular momentum of the system about this axis is conserved.

   The principle of conservation of angular momentum states: $L_{initial} = L_{final}$ where $L = I \omega$ is the angular momentum, $I$ is the moment of inertia, and $\omega$ is the angular speed.

2. Analyze the Initial State

   • The ring has mass $M$ and radius $R$. Its moment of inertia about the central axis is $I_{ring} = MR^2$.
   • Initially, two point masses, each of mass $m = \frac{M}{8}$, are at the center O. Their distance from the axis of rotation is $r=0$.
   • The initial moment of inertia of the two masses is $I_{masses, i} = m(0)^2 + m(0)^2 = 0$.
   • The total initial moment of inertia of the system is: $I_i = I_{ring} + I_{masses, i} = MR^2 + 0 = MR^2$
   • The initial angular speed is $\omega_i = \omega$.
   • The initial angular momentum of the system is: $L_i = I_i \omega_i = (MR^2) \omega$

3. Analyze the Final State

   • At the instant in question, the ring is still rotating, so its moment of inertia remains $I_{ring} = MR^2$.
   • One mass ($m_1 = \frac{M}{8}$) is at a distance $r_1 = \frac{3}{5}R$ from O. Its moment of inertia is: $I_1 = m_1 r_1^2 = \left(\frac{M}{8}\right) \left(\frac{3}{5}R\right)^2 = \frac{M}{8} \frac{9R^2}{25} = \frac{9}{200} MR^2$
   • The other mass ($m_2 = \frac{M}{8}$) is at an unknown distance $r_2$ from O. Its moment of inertia is: $I_2 = m_2 r_2^2 = \frac{M}{8} r_2^2$
   • The total final moment of inertia of the system is: $I_f = I_{ring} + I_1 + I_2 = MR^2 + \frac{9}{200} MR^2 + \frac{M}{8} r_2^2$
   • The final angular speed is given as $\omega_f = \frac{8}{9}\omega$.
   • The final angular momentum of the system is: $L_f = I_f \omega_f = \left( MR^2 + \frac{9}{200} MR^2 + \frac{M}{8} r_2^2 \right) \left( \frac{8}{9} \omega \right)$

4. Apply Conservation of Angular Momentum Equating the initial and final angular momentum ($L_i = L_f$): $MR^2 \omega = \left( MR^2 + \frac{9}{200} MR^2 + \frac{M}{8} r_2^2 \right) \left( \frac{8}{9} \omega \right)$

5. Solve for the unknown distance $r_2$

   • We can cancel $\omega$ from both sides (as $\omega \neq 0$). $MR^2 = \left( MR^2 + \frac{9}{200} MR^2 + \frac{M}{8} r_2^2 \right) \frac{8}{9}$
   • Multiply both sides by $\frac{9}{8}$: $\frac{9}{8} MR^2 = MR^2 + \frac{9}{200} MR^2 + \frac{M}{8} r_2^2$
   • Move the terms with $MR^2$ to the left side: $\frac{9}{8} MR^2 - MR^2 - \frac{9}{200} MR^2 = \frac{M}{8} r_2^2$
   • Factor out $MR^2$ on the left side: $MR^2 \left( \frac{9}{8} - 1 - \frac{9}{200} \right) = \frac{M}{8} r_2^2$
   • Simplify the expression in the parenthesis: $\frac{9}{8} - 1 = \frac{1}{8}$ $MR^2 \left( \frac{1}{8} - \frac{9}{200} \right) = \frac{M}{8} r_2^2$
   • Find a common denominator for 8 and 200, which is 200: $MR^2 \left( \frac{25}{200} - \frac{9}{200} \right) = \frac{M}{8} r_2^2$ $MR^2 \left( \frac{16}{200} \right) = \frac{M}{8} r_2^2$
   • Cancel $M$ from both sides: $R^2 \frac{16}{200} = \frac{r_2^2}{8}$
   • Simplify the fraction $\frac{16}{200}$ by dividing numerator and denominator by 8: $R^2 \frac{2}{25} = \frac{r_2^2}{8}$
   • Solve for $r_2^2$: $r_2^2 = 8 \times \frac{2}{25} R^2 = \frac{16}{25} R^2$
   • Take the square root of both sides (distance must be positive): $r_2 = \sqrt{\frac{16}{25} R^2} = \frac{4}{5} R$

6. Conclusion The distance of the other mass from O is $\frac{4}{5}R$. This corresponds to option D.