1. Moment of inertia of a disc about its central axis

For a solid circular disc, the moment of inertia about an axis through its center and perpendicular to its plane is

$I_{CM} = \frac{1}{2}MR^2$

where $M$ is the mass of the disc and $R$ is its radius.

2. Moment of inertia about parallel axis $AB$

The axis $AB$ is parallel to the central axis and is at a distance

$d = \frac{2R}{3}$

from the center.

By the parallel axis theorem,

$I_{AB} = I_{CM} + Md^2$

Substitute the values:

$I_{AB} = \frac{1}{2}MR^2 + M\left(\frac{2R}{3}\right)^2$

$I_{AB} = \frac{1}{2}MR^2 + M\cdot \frac{4R^2}{9}$

$I_{AB} = MR^2\left(\frac{1}{2} + \frac{4}{9}\right)$

Take LCM $18$:

$I_{AB} = MR^2\left(\frac{9+8}{18}\right) = MR^2\cdot \frac{17}{18}$

3. Find the ratio $I_{AB} : I_{CM}$

$\frac{I_{AB}}{I_{CM}} = \frac{\frac{17}{18}MR^2}{\frac{1}{2}MR^2}$

Cancel $MR^2$:

$\frac{I_{AB}}{I_{CM}} = \frac{17}{18} \cdot 2 = \frac{17}{9}$

So,

$I_{AB} : I_{CM} = 17 : 9$

Comparing with $x:9$, we get

$x=17$

4. Comparison with stored answer

Stored correct answer = $17$

This matches the derived answer.