= Solution
Measure $z$ from the apex along the cone's symmetry axis. A thin disc at height $z$ has radius
$$
s(z)=\frac RH z
$$
and mass $dM=\rho\pi s^2\,dz$. Its moment about a parallel diameter through its centre is $s^2dM/4$, while its centre is distance $z$ from the required axis. Part (a)(ii) gives
$$
dI=\left(z^2+\frac{s^2}{4}\right)dM.
$$
Hence
$$
\begin{aligned}
I
&=\rho\pi\int_0^H
\frac{R^2z^2}{H^2}
\left(z^2+\frac{R^2z^2}{4H^2}\right)dz\\
&=\rho\pi\left(\frac{R^2H^3}{5}
+\frac{R^4H}{20}\right).
\end{aligned}
$$
Since $M=\rho\pi R^2H/3$,
$$
\boxed{
I=M\left(\frac3{20}R^2+\frac35H^2\right)},
\qquad
\boxed{\alpha=\frac3{20},\quad\beta=\frac35}.
$$
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