= Solution
For the <edge-hinged cube with a central spherical cavity>, use the parallel-axis theorem: for a parallel axis at perpendicular distance $d$ from the centre-of-mass axis, $I=I_C+Md^2$. Let $M=\rho a^3$. The centre lies $d=a/\sqrt2$ from the edge hinge, so
$$
\boxed{I_h=Ma^2/6+M(a/\sqrt2)^2=\frac23Ma^2=\frac23\rho a^5.}
$$
At equilibrium the centre is vertically below the hinge. Its gravitational <torque> at angular displacement $\theta$ is $-Mgd\sin\theta$. Therefore the small-angle equation is $I_h\ddot\theta+Mgd\theta=0$, with
$$
\boxed{\omega^2=\frac{Mgd}{I_h}=\frac{3g}{2\sqrt2a}.}
$$
The removed <sphere> has <mass> $M_s=\pi M/6$ and the same centre. Its hinge inertia is
$$
I_{s,h}=M_s\left[\frac25(a/2)^2+(a/\sqrt2)^2\right]=\frac35M_sa^2=\frac\pi{10}Ma^2.
$$
<Superposition> leaves the centre at the same location but changes both inertia and weight:
$$
\boxed{I_{\rm hollow}=Ma^2\left(\frac23-\frac\pi{10}\right),\qquad
\Omega=\omega\sqrt{\frac{1-\pi/6}{1-3\pi/20}}\approx0.9492\omega.}
$$
The <frequency> decreases: the removed central material contributes less hinge inertia per unit <mass> than the original cube average.
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