That's why I like physics

The points where the ring segments meet are distance R apart, so the angle of reach segment is 60° or $\frac{π}{3}$ radians.

Moment of inertia $L=\int r^2dm$ .

Note:

• we count 4 half segments, each with angle running from $0$ to $\frac{π}{6}$
• r is the distance to the axis of rotation (the line AB), we express it as a function of the angle: $r=R(\cos φ-\cos\frac{π}{6})$ .
• the mass per angle then is $\frac{dm}{dφ} = \frac{3M}{π}$

$L = 4\int_0^{\frac{π}{6}}R^2(\cos φ-\cos \frac{π}{6})^2 \frac{3M}{π}dφ$

$= \frac{12R^2M}{π}\int_0^{\frac{π}{6}} \cos^2 φ- \sqrt{3}\cos φ +\frac{3}{4} dφ$

$= \frac{12R^2M}{π}\int_0^{\frac{π}{6}} \frac{1}{2}\cos 2φ + \frac{1}{2} - \sqrt{3}\cos φ +\frac{3}{4} dφ$

$= \frac{12R^2M}{π} (\frac{1}{4}\sin \frac{π}{3} + \frac{π}{12} - \sqrt{3} \sin\frac{π}{6} +\frac{3}{4} \frac{π}{6})$

$= \frac{12R^2M}{π} (\frac{5π}{24}-\frac{3}{8}\sqrt{3})$

$= \frac{R^2M}{2} (5  -\frac{9\sqrt{3}}{π})$

$5+9=\boxed{14}$

@Ayush Choubey exactly

Sorry , I would not be able to post the whole solution but the sol. will surely answer the question asked

In the given fig . "c" is centre of mass of the portion of the ring. O is centre of the ring and we need to calculate "I(P)"

You can calculate it using the method to calculate that of a semicircular ring and it comes out to be - $\frac { 3R }{ \pi }$ = OC

I (O) - M(OC )^2 = I(C.O.M) ... Using parallel axis theorem .

I (P) = I (C.O.M) + M(CP)^2

where cp = oc - op = $\frac { 3R }{ \pi }$ - $\sqrt { 3 } \frac { R }{ 2 }$

( OP is calculated using trignometry)

Now , since both portions are similar ,

Net Moment of Inertia = 2 * I(P)

giving c = 5 and d = 9

Answer = $\boxed {14}$

I apologize if something is missing or unclear .