Answer
$\theta=tan^{-1}(\frac{2}{5\mu})$
Work Step by Step
Please see the attached image first.
We'll get the minimum angle when static friction is greatest. At that time $F_{S}=\mu n_{1}$
Let's consider the static equilibrium of the board.
$\uparrow \sum \vec F=0$
$n_{1}-mg=0$
$n_{1}=mg-(1)$
$\rightarrow \sum\vec F=0$
$F_{S}-n_{1}=0$
$\mu n_{1}=n_{2}-(2)$
$(1)=\gt (2)$
$\mu mg=n_{2}-(3)$
$\sum \tau_{A}\curvearrowright=0$
$m\times g\frac{2L}{5}cos\theta -n_{2}Lsin\theta=0$
$(3)=\gt$
$\frac{2mg}{5}cos\theta -\mu mgsin\theta=0$
$tan\theta=\frac{2}{5\mu}=\gt \theta=tan^{-1}(\frac{2}{5\mu})$
