A given object takes $n$  times as much time to slide down a $45°$  incline if the friction is present.  The coefficient of friction between the object and the incline is 

For smooth surface,   $s=\frac{1}{2}g\sin \theta t_1^2\mathrm{....}\left(i\right)$
For rough surface,        $a=g(\sin \theta -\mu \cos \theta )t_2^2$  
 $\therefore s=\frac{1}{2}g(\sin \theta -\mu \cos \theta )t_2^2\mathrm{...}\left(ii\right)$
From (i) and (ii),
 $\therefore s=\frac{1}{2}g\sin \theta t_1^2=\frac{1}{2}g(\sin \theta -\mu \cos \theta )t_2^2$
Given,    $\theta =45°\therefore t_1^2=(1-\mu )t_2^2$
Also, given that,   $t_2=n{t}_1\therefore t_1^2=(1-\mu )n^2{t}_1^2$
 $\frac{1}{n^2}=1-\mu \therefore \mu =(1-\frac{1}{n^2})$