Answer
$F=\frac{2\mu mg}{1-\mu}$
Work Step by Step
We know that
$\Sigma \tau=0$
$\implies (Tcos45^{\circ})L-F(\frac{L}{2})=0$
This simplifies to:
$\frac{T}{\sqrt 2}=\frac{F}{2}$.....eq(1)
Similarly $\Sigma F_y=0$
$\implies N-Tsin45-mg=0$
This simplifies to:
$N=\frac{T}{\sqrt 2}+mg$.....eq(2)
Now $\Sigma F_x=0$
$\implies -Tcos45-f+F=0$
$-\frac{T}{\sqrt 2}-\mu N+F=0$
$\implies -\frac{T}{\sqrt 2}-\mu (\frac{T}{\sqrt 2}+mg)+F=0$
$\implies \frac{F}{2}=\mu (\frac{F}{2}+mg)$
This simplifies to:
$F=\frac{2\mu mg}{1-\mu}$
