 Solution

For a uniformly distributed pure qubit on the Bloch sphere, rotational symmetry gives $⟨r_z^2⟩=1/3$. ThusEquivalently, $p=|\alpha {|}^2$ is uniform on $[0,1]$ under this measure and ${\int }_0^1[p^2+(1-p{)}^2]dp=2/3$. The three positive coordinate-axis states mentioned in the source give $F_Z=1$, $F_X=F_Y=1/2$, whose average is the same because this fidelity depends only on the quadratic component $r_z^2$. This is a specified uniform-state average, not a statement about every possible prior ensemble.