= Solution
For the <moment of inertia of a uniform solid sphere>, write the <mass density> as $\rho=3m/(4\pi a^3)$. In spherical coordinates about the chosen axis, the squared perpendicular distance is $R^2\sin^2\vartheta$. The volume element is $R^2\sin\vartheta\,dR\,d\vartheta\,d\varphi$, so
$$
I=\rho\int_0^aR^4\,dR\int_0^\pi\sin^3\vartheta\,d\vartheta
\int_0^{2\pi}d\varphi
=\rho\frac{a^5}{5}\frac43\,2\pi
=\boxed{\frac25ma^2}.
$$
Take $0<\alpha<\pi/2$ and $\ell>0$, with a fixed plane. The <normal force> has magnitude $N=mg\cos\alpha$ and is perpendicular to the plane. On the perfectly smooth plane there is no friction. Both gravity and the <normal force> have zero <torque> about the marble's centre, so a marble released from rest does not rotate. The component of <Newton's second law> down the plane is $mA_1=mg\sin\alpha$, hence
$$
A_1=g\sin\alpha,\qquad
\boxed{t_1=\sqrt{\frac{2\ell}{g\sin\alpha}}}.
$$
The <normal force> does no work because the velocity is tangent to the plane. Gravity is conservative, so <mechanical energy>, the sum of translational <kinetic energy> and <gravitational potential energy>, is conserved.
For <rolling without slipping>, let the <static friction> force have magnitude $f$, pointing up the plane. It supplies the <torque> that spins the marble while reducing its translational acceleration. If $\Omega$ is the <angular velocity> and $A_2$ the acceleration down the plane, the rolling constraint is $a\Omega=\dot s$, hence $a\dot\Omega=A_2$. The translation and rotation equations are
$$
mA_2=mg\sin\alpha-f,\qquad
fa=I\dot\Omega=\frac{I A_2}{a}.
$$
Thus $f=IA_2/a^2$ and
$$
A_2=\frac{g\sin\alpha}{1+I/(ma^2)}=\frac57g\sin\alpha,
\qquad f=\frac27mg\sin\alpha.
$$
These are the equations for <rolling acceleration with rotational inertia>; the plane must be rough enough to supply this <static friction>.
The point of contact is instantaneously at rest relative to the fixed plane, so <static friction> does no total work on the rolling marble. Equivalently, its translational power is $-f\dot s$ and its rotational power is $fa\Omega=f\dot s$, which cancel. The <normal force> also does no work. Consequently <mechanical energy> is conserved here too, now with <rotational kinetic energy> included:
$$
\frac12m\dot s^2+\frac12I\Omega^2+mgH=\text{constant},
$$
where $H$ is the centre's height. No energy loss is implied by the presence of ideal <static friction>. Constant acceleration from rest gives
$$
\boxed{t_2=\sqrt{\frac{14\ell}{5g\sin\alpha}},\qquad
\frac{t_1}{t_2}=\sqrt{\frac57}.}
$$
For the hollow marble, put $I=\lambda ma^2$ in the same equations. The smooth-plane time is unchanged, while the rolling acceleration becomes $g\sin\alpha/(1+\lambda)$. Therefore
$$
\boxed{t_2=\sqrt{\frac{2\ell(1+\lambda)}{g\sin\alpha}},\qquad
\frac{t_1}{t_2}=\frac1{\sqrt{1+\lambda}}.}
$$
This again assumes <rolling without slipping> and enough <static friction>; the ratio depends on the mass distribution through its <moment of inertia>.
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