Answer
$\tau_{y}= \sigma \sin \theta \cos \theta$
Work Step by Step
In the y-direction, the force will be
$$F_{y}=F \sin \theta$$
Hence, to get the shear stress at the static equilibrium, it will be
$$\tau_{y}=\frac{F^{y}}{A^{y}}=\frac{F \sin \theta}{\dfrac{A}{\cos \theta}}=\frac{F}{A} \sin \theta \cos \theta=\boxed{\sigma \sin \theta \cos \theta}$$
