Let length is $\ell$ of inclined plane, then $f_r=\mu N =\mu mg \cos \theta$

$mg \sin \theta- f _{ r }= ma$

$mg \sin \theta-\mu mg \cos \theta= ma$

Now

$\ell=\frac{1}{2} at ^2=\frac{1}{2} g (\sin \theta-\mu \cos \theta) t ^2$

$\text { Now } \ell_1=\ell_2$

$\ell_2=\frac{1}{2} g \sin \theta\left(\frac{ t ^2}{2}\right)$

$\frac{1}{2} g(\sin \theta-\mu \cos \theta) t ^2=\frac{1}{2} g(\sin \theta-0 \times \cos \theta)\left(\frac{ t ^2}{2}\right)$

$4(\sin \theta-\mu \cos \theta)=\sin \theta$

$\mu=\frac{3}{4}=0.75$