If disk slips on inclined plane, then its acceleration, $a_1=g\sin \mathrm{\theta }$
$L=\frac{1}{2}{a}_1{t}_1^2$
$\Rightarrow t_1=\sqrt{\frac{2 L}{a_1}} \dots \left(\mathrm{i}\right)$

If disk rolls on inclined plane, its acceleration,
$a_2=\frac{g\sin \mathrm{\theta }}{1+\frac{I}{m{R}^2}}$
$a_2=\frac{g\sin \mathrm{\theta }}{1+\frac{m{R}^2}{2 m{R}^2}}$
$a_2=\frac{2}{3}g\sin \mathrm{\theta }$
Now $L=\frac{1}{2}{a}_2·t_2^2$
$\Rightarrow t_2=\sqrt{\frac{2 L}{a_2}} \dots \left(\mathrm{ii}\right)$

Now, $\frac{t_2}{t_1}=\sqrt{\frac{a_1}{a_2}}=\sqrt{\frac{3}{2}}$
$\Rightarrow x=2$