Answer
$\theta$ must be at least $\arctan\frac{1}{(2\mu_s)}$
Work Step by Step
We sum the forces vertically to relate the normal force to the mass of the ladder:
$(\uparrow +) \sum \overrightarrow{F}_{y} =0 $
$-mg+F_N=0$
$F_N=mg$
Then, we sum the torques about the top of the ladder and solve for $\theta$:
$(+ \circlearrowleft) \sum \tau=0$
$-\cos\theta*F_N*l-\sin\theta*\mu_s*F_N*l+mg\cos\theta*1/2*l=0$
$-\cos\theta*F_N-\sin\theta*\mu_s*F_N+mg\cos\theta*1/2=0$
$-\cos\theta*F_N-\sin\theta*\mu_s*F_N+mg\cos\theta*1/2=0$
$-\cos\theta*F_N-\sin\theta*\mu_s*F_N+mg\cos\theta*1/2=0$
$\cos\theta(-F_N+mg*1/2)-\sin\theta*\mu_s*F_N=0$
$\cos\theta(-F_N+mg*1/2)=\sin\theta*\mu_s*F_N$
$\frac{-F_N+mg*1/2)}{\mu_s*F_N}=\frac{\sin\theta}{\cos\theta}$
$\frac{-mg+mg*1/2)}{\mu_s*mg}=\tan\theta$
$\frac{-1+1/2}{\mu_s}=\tan\theta$
$\frac{1}{2\mu_s}=\tan\theta$
$\theta=\arctan1/(2\mu_s)$
Therefore, $\theta$ must be at least $\arctan\frac{1}{(2\mu_s)}$.