= Solution
At height $z$ from the vertex, the cone has cross-sectional area $\pi R^2z^2/l^2$. Hence
$$
z_{\rm CM}=\frac{\int_0^l z^3\,dz}{\int_0^l z^2\,dz}=\frac{3l}{4},
\qquad
\mathbf r_{\rm CM}=a\mathbf e_3,
\quad a=\frac{3l}{4}.
$$
The parallel-axis theorem is
$$
\mathbf I_O=\mathbf I_{\rm CM}
+M\{|mathbf a|^2\mathbf1-\mathbf a\mathbf a^T\}.
$$
The displacement is along $\mathbf e_3$, so it changes the two transverse moments but not the axial one. Thus, writing the vertex moments as $I_1=I_2=I$ and $I_3=J$,
$$
\boxed{I=I_1^{\rm CM}+Ma^2
=\frac{\pi R^2l\rho}{20}(R^2+4l^2)
=\frac{3M}{20}(R^2+4l^2)},
$$
and
$$
\boxed{J=I_3^{\rm CM}
=\frac{\pi R^4l\rho}{10}
=\frac{3MR^2}{10}}.
$$
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