We know that the force acting on the area element $$dS$$ of a conductor is,
$$d\vec {F} = \dfrac {1}{2} \sigma \vec {E} dS .... (1)$$
It follows from symmetry considerations that the resultant force $$\vec {F}$$ is directed along the $$z-axis$$, and hence it can be represented as the sum (integral) of the projection of elementary forces (1) onto the z-axis:
$$dF_{z} = dF \cos \theta .... (2)$$
For simplicity let us consider an element area $$dS = 2\pi R \sin \theta R d\theta$$ (Fig.). Now considering that $$E = \sigma/ \epsilon_{0}$$. Equation (2) takes the from
$$dF_{z} = \dfrac {\pi \sigma^{2}R^{2}}{\epsilon_{0}} \sin \theta \cos \theta d\theta$$
$$= -\left (\dfrac {\pi \sigma_{0}^{2} R^{2}}{\epsilon_{0}}\right ) \cos^{3}\theta d\cos \theta$$
Integrating this expression over the half sphere (i.e. with respect to $$\cos \theta$$ between $$1$$ and $$0$$), we obtain
$$F = F_{z} = \dfrac {\pi \sigma_{0}^{2} R^{2}}{4\epsilon_{0}}$$.