Let angle of inclination is θ=45∘.
Then, acceleration along the incline in case of smooth incline will be, a1=gsinθ=2g.
In case of rough incline, kinetic friction will act against the direction of motion. Normal force acting on the object will be mgcosθ. Therefore, acceleration along the incline in case of rough incline will be,
a2=gsinθ−μgcosθ=2g−2μg.
Now as per question,
t2=nt1
and as the displacement covered is the same in both case, 0+21a1t12=0+21a2t22
⇒2gt12=(2g−2μg)n2t12
⇒μ=1−n21