= Solution
Take the upper hemisphere to have $z>0$. Integrate the force $\mathbf E\,dq$ using the total interior <electric field> from part (a). This computes the net external force on the upper hemisphere, since the forces between pairs of its own charge elements cancel by antisymmetry of the Coulomb force. Horizontal components cancel by symmetry. The vertical component is
$$
F_z=\frac{\rho_e^2}{3\varepsilon_0}\int_{z>0,\,r<R}z\,dV.
$$
A horizontal section at height $z$ has area $\pi(R^2-z^2)$, so
$$
\int z\,dV=\pi\int_0^R z(R^2-z^2)\,dz=\frac{\pi R^4}{4}.
$$
Thus \b[the lower hemisphere repels the upper one vertically upward], with
$$
\boxed{\mathbf F=\frac{\pi\rho_e^2R^4}{12\varepsilon_0}\hat{\mathbf z}=\frac{3Q^2}{64\pi\varepsilon_0R^2}\hat{\mathbf z}.}
$$
The result is the same for either sign of $Q$, as expected for the <force between hemispheres of a uniformly charged ball>.
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