$s=ut+\frac{1}{2}a{t}^2,a=g\sin \theta ,u=0$

Then, $s=0+\frac{1}{2}a{t}^2$

or $t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2s}{g\sin \theta }}$ [for smooth plane]

For the rough plane. the effective value of acceleration along the incline is
$a^{\text{'}}=g\sin \theta -{\mu }_{k}g\cos \theta$

and $t^{\text{'}}=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2s}{g\sin \theta -{\mu }_{k}g\cos \theta }}$

Now, $\frac{t^{\text{'}}}{t}=n=\sqrt{\frac{g\sin \theta }{g\sin \theta -{\mu }_{k}g\cos \theta }}$

$=\sqrt{\frac{\sin 45°}{\sin 45°-{\mu }_k\cos 45°}}$

$=\sqrt{\frac{1}{1-{\mu }_k}}$

$\Rightarrow \frac{1}{n^2}=1-{\mu }_k\Rightarrow {\mu }_k=1-\frac{1}{n^2}$