I _ CM is moment of inertia of a circular disc about an axis (CM)...

We have, ${I}_{CM} = \dfrac{1}{2}M{R}^{2}$

Since $AB$ is parallel to diameter and at a distance of $\dfrac{2}{3}R$ from it, using parallel axis theorem,

${I}_{AB} = \dfrac{m{R}^{2}}{2} + m{\left(\dfrac{2R}{3}\right)}^{2} = \dfrac{17}{18}m{R}^{2}$

Thus, the ratio is

$∴ \dfrac{{I}_{AB}}{{I}_{CM}} = \dfrac{\dfrac{17}{18}M{R}^{2}}{\dfrac{1}{2}M{R}^{2}} = \dfrac{17}{9}$

Thus, $x = 17$