The forces acting on block is shown in above figure.

Here, normal force, $N=mg\cos 45°$ and frictional force, $f=\mu N$.

Balancing the force along incline, we have

$F_1=mg \sin 45°+f=mg \sin 45°+\mu N$

$\Rightarrow F_1=\frac{mg}{\sqrt{2}}+\mu mg \cos 45°$

$\Rightarrow F_1=\frac{mg}{\sqrt{2}}\left(1+\mu \right)$

Now, the forces acting on block to just push it up on inclined plane is shown below.

Balancing the force along incline, we have

$F_2=mg \sin 45°-f=mg \sin 45°-\mu N$

$=\frac{mg}{\sqrt{2}}\left(1-\mu \right)$

Given, $F_1=2F_2$

$\Rightarrow \frac{mg}{\sqrt{2}}\left(1+\mu \right)=2\frac{mg}{\sqrt{2}}\left(1-\mu \right)$

$\Rightarrow 1+\mu =2-2\mu$

$\Rightarrow \mu =\frac{1}{3}=0.33$