= Solution
Write the position of a mass element relative to the <center of mass> as $\mathbf r$, so its inertial-frame velocity is $\dot{\mathbf X}+\boldsymbol\omega\times\mathbf r$. The center-of-mass identity $\int\mathbf r\,dm=0$ removes all cross terms. Define the <inertia tensor> about the center of mass by
$$
I_{ij}=\int\bigl(r^2\delta_{ij}-r_ir_j\bigr)\,dm.
$$
The total <angular momentum> about the origin and <kinetic energy> are then
$$
\mathbf L=M\mathbf X\times\dot{\mathbf X}+I\boldsymbol\omega,
\qquad
T=\frac12M|\dot{\mathbf X}|^2+\frac12\boldsymbol\omega\mathbin{\cdot}I\boldsymbol\omega.
$$
Let $\rho(r)=C r^{-1}\sin(\pi r/R)$. <Spherical symmetry> makes $I=I_0\mathbf1$. The mass normalization is
$$
M=4\pi C\int_0^R r\sin\frac{\pi r}{R}\,dr=4CR^2,
$$
so $C=M/(4R^2)$. A moment about any diameter is
$$
I_0=\frac{8\pi}{3}\int_0^R\rho(r)r^4\,dr
=\frac{8\pi C}{3}\frac{R^4}{\pi^4}\int_0^\pi x^3\sin x\,dx
=\frac{2MR^2(\pi^2-6)}{3\pi^2}.
$$
Therefore
$$
\boxed{I=\frac{2MR^2(\pi^2-6)}{3\pi^2}\,\mathbf1}.
$$
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