Moment of inertia about the diameter of the circular loop (ring) $$ = \frac{1}{2}M{R^2}$$
Using parallel axis theorem
The moment of inertia of the loop about $$XX’$$  axis is
$${I_{XX'}} = \frac{{M{R^2}}}{2} + M{R^2} = \frac{3}{2}M{R^2}$$
Where $$M\,=$$  mass of the loop and $$R \,=$$  radius of the loop
Here $$M = L\rho $$   and $$R = \frac{L}{{2\pi }};$$
$$\therefore {I_{XX'}} = \frac{3}{2}\left( {L\rho } \right){\left( {\frac{L}{{2\pi }}} \right)^2} = \frac{{3{L^3}\rho }}{{8{\pi ^2}}}$$

Considering the information given in the question, let us draw the figure

If the above figure is considered, then moment of inertia of disc will be given as
$$I = {I_{{\text{remain}}}} + {I_{\left( {\frac{R}{2}} \right)}} \Rightarrow {I_{{\text{remain}}}} = I - {I_{\left( {\frac{R}{2}} \right)}}$$
Putting the values, we get
$$\eqalign{ & = \frac{{M{R^2}}}{2} - \left[ {\frac{{\frac{M}{4}{{\left( {\frac{R}{2}} \right)}^2}}}{2} + \frac{M}{4}{{\left( {\frac{R}{2}} \right)}^2}} \right] \cr & = \frac{{M{R^2}}}{2} - \left[ {\frac{{M{R^2}}}{{32}} + \frac{{M{R^2}}}{{16}}} \right] \cr & = \frac{{M{R^2}}}{2} - \left[ {\frac{{M{R^2} + 2M{R^2}}}{{32}}} \right] \cr & = \frac{{M{R^2}}}{2} - \frac{{3M{R^2}}}{{32}} = \frac{{16M{R^2} - 3M{R^2}}}{{32}} \cr & {I_{{\text{remain }}}} = \frac{{13M{R^2}}}{{32}} \cr} $$

As we know that
Angular momentum $$L = m\left( {r \times v} \right)$$
So, here angular momentum is directed along a line perpendicular to the plane of rotation.

According to parallel axis theorem of the moment of Inertia
$$I = {I_{cm}} + m{d^2}$$
$$d$$ is maximum for point $$B$$ so $${I_{\max }}$$  about $$B.$$

$$\eqalign{ & {\text{Here }}a = \frac{2}{{\sqrt 3 }}R \cr & {\text{Now, }}\frac{M}{{M'}} = \frac{{\frac{4}{3}\pi {R^3}}}{{{a^3}}} \cr & = \frac{{\frac{4}{3}\pi {R^3}}}{{{{\left( {\frac{2}{{\sqrt 3 }}R} \right)}^3}}} \cr & = \frac{{\sqrt 3 }}{2}\pi .\,\,\,\,\,\,\,\,\,\,M' = \frac{{2m}}{{\sqrt 3 \pi }} \cr} $$
Moment of inertia of the cube about the given axis,
$$I = \frac{{M'{a^2}}}{6} = \frac{{\frac{{2M}}{{\sqrt 3 \pi }} \times {{\left( {\frac{2}{{\sqrt 3 }}R} \right)}^2}}}{6} = \frac{{4M{R^2}}}{{9\sqrt 3 \pi}}$$

The moment of inertia of a body about an axis depends not only on the mass of the body, but also on the distribution of mass from the axis. For a given body, mass is same, so it will depend only on the distribution of mass from the axis.
The mass is farthest from axis $$BC,$$  so $${I_2}$$ is maximum. Mass is nearest to axis $$AC,$$  so $${I_3}$$ is minimum. Hence, the correct sequence will be $${I_2} > {I_1} > {I_3}$$
NOTE
In a rotational motion, moment of inertia is also known as rotational inertia.

For translational motion,
$$mg-T=ma\,.....(1)$$
For rotational motion,
$$\eqalign{ & {\text{T}}{\text{.R}}{\text{.}} = I\alpha = I\frac{a}{R}\,.....(2) \cr & {\text{Solving (1) and (2)}} \cr & a = \frac{{mg}}{{\left( {m + \frac{I}{{{R^2}}}} \right)}} = \frac{{mg}}{{m + \frac{{m{R^2}}}{{2{R^2}}}}} = \frac{{2mg}}{{3m}} = \frac{{2g}}{3} \cr} $$

Torque on the rod is equal to moment of weight of rod about $$P.$$

Torque on the rod = Moment of weight of the rod about $$P$$
$$\tau = Mg\frac{L}{2}\,......\left( {\text{i}} \right)$$
$$\because $$ Moment of inertia of rod
about $$P = \frac{{M{L^2}}}{3}\,......\left( {{\text{ii}}} \right)$$
As $$\tau = I\alpha $$
From Eqs. (i) and (ii), we get
$$Mg\frac{L}{2} = \frac{{M{L^2}}}{3}\alpha \Rightarrow \alpha = \frac{{3g}}{{2L}}$$

$${I_{xy}},$$  moment of inertia of a ring about its tangent in the plane of ring $${I_{{x^1}y}} = \frac{3}{2}M{R^2}$$
Moment of inertia about a tangent perpendicular to the plane of ring $${I_{xy}} = 2M{R^2}$$
$$\therefore {I_{xy}} = \frac{3}{4}\left( {2M{R^2}} \right) = \frac{3}{2}M{R^2}\,or\,{I_{xy}} = \frac{3}{4}{I_{{x^1}{y^1}}}$$

Since there is no external torque, angular momentum remains conserved. As moment of inertia initially decreases and then increases, so $$\omega $$  will increase initially and then decreases.

Note : The $$M.I.$$  of the system decreases when the tortoise move from $$A$$  to $$B$$  and then increases from $$B$$  to $$A.$$
So the variation of $$\omega $$  is nonlinear.