Consider a block kept on an inclined plane (inclined at 45 ° ) as...

The forces acting on block is shown in above figure.

Here, normal force, $N = mgcos45 °$ and frictional force, $f = \muN$.

Balancing the force along incline, we have

${F}_{1} = mgsin45 ° + f = mgsin45 ° + \muN$

$⇒ {F}_{1} = \dfrac{mg}{\sqrt{2}} + \mu mgcos45 °$

$⇒ {F}_{1} = \dfrac{mg}{\sqrt{2}}\left(1 + \mu\right)$

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

Balancing the force along incline, we have

${F}_{2} = mgsin45 ° - f = mgsin45 ° - \muN$

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

Given, ${F}_{1} = 2{F}_{2}$

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

$⇒ 1 + \mu = 2 - 2\mu$

$⇒ \mu = \dfrac{1}{3} = 0 . 33$