
The forces acting on block is shown in above figure.
Here, normal force, N=mgcos45∘ and frictional force, f=μN.
Balancing the force along incline, we have
F1=mgsin45∘+f=mgsin45∘+μN
⇒F1=2mg+μmgcos45∘
⇒F1=2mg(1+μ)
Now, the forces acting on block to just push it up on inclined plane is shown below.

Balancing the force along incline, we have
F2=mgsin45∘−f=mgsin45∘−μN
=2mg(1−μ)
Given, F1=2F2
⇒2mg(1+μ)=22mg(1−μ)
⇒1+μ=2−2μ
⇒μ=31=0.33