A circular disc reaches from top to bottom of an inclined plane of...

If disk slips on inclined plane, then its acceleration, ${a}_{1} = gsin\theta$
$L = \dfrac{1}{2}{a}_{1}{t}_{1}^{2}$
$⇒ {t}_{1} = \sqrt{\dfrac{2L}{{a}_{1}}} \ldots \left(i\right)$

If disk rolls on inclined plane, its acceleration,
${a}_{2} = \dfrac{gsin\theta}{1 + \dfrac{I}{m{R}^{2}}}$
${a}_{2} = \dfrac{gsin\theta}{1 + \dfrac{m{R}^{2}}{2m{R}^{2}}}$
${a}_{2} = \dfrac{2}{3}gsin\theta$
Now $L = \dfrac{1}{2}{a}_{2} \cdot {t}_{2}^{2}$
$⇒ {t}_{2} = \sqrt{\dfrac{2L}{{a}_{2}}} \ldots \left(ii\right)$

Now, $\dfrac{{t}_{2}}{{t}_{1}} = \sqrt{\dfrac{{a}_{1}}{{a}_{2}}} = \sqrt{\dfrac{3}{2}}$
$⇒ x = 2$