= Solution
Using spherical shells and averaging the squared distance from the axis gives
$$
I=\frac25Ma^2.
$$
Sliding without friction has <acceleration> $g\sin\alpha$, so
$$
t_s=\sqrt{\frac{2l}{g\sin\alpha}}.
$$
For rolling, $Ma_{\rm cm}=Mg\sin\alpha-F$ and $I(a_{\rm cm}/a)=Fa$, whence
$$
a_{\rm cm}=\frac{g\sin\alpha}{1+I/(Ma^2)}.
$$
The <rolling acceleration with rotational inertia> gives $5g\sin\alpha/7$ for the uniform sphere, and therefore
$$
\frac{t_s}{t_r}=\sqrt{\frac57}.
$$
Mechanical energy is conserved in both idealizations: there is no friction in sliding, while static friction does no work at the instantaneous contact point in pure rolling.
For $I=\gamma Ma^2$, the same calculation gives
$$
\boxed{\frac{t_s}{t_r}=\frac1{\sqrt{1+\gamma}}.}
$$
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