Conceptual Assessment of the Situation
First, in order to provide rotation (which it will) the force needs to move a bit with the surface. Otherwise the only energy transferred by the force will be along the distance of total translation, which cannot account for the work energy that created the angular kinetic energy.
At first blush, it seems this could be assumed to be from perfectly elastic collisions (elastic meaning no loss of energy during collision, and hence no sound energy or heat energy generated, and the net energy transfer from the change in velocity - magnitude that is - of the particles) from infinitely-many infinitesimal objects or a fluid jet. However, that doesnt work because it’s impossible to apply a tangential force that way.
One somewhat reasonable view of how this is happening is that a small charged bead is in the insulating ball just inside its surface, and the whole experiment is in a vertical electric field. The bead can move around the circumference freely. The problem is the dynamics of the bead transferring momentum to the ball. The real-world situation would be close.
That said, as with many such, the problem is theoretical (and technically theoretically impossible).
Concept of the Model
Mechanism aside, one important concept is that there can be no rotational energy imparted if the force only moves with the ball, ie as much as (and along a parallel path with) the movement of the center of mass of the ball. But we see due to the radial offset of the force from center that angular kinetic energy will develop.
Therefore, the conceptual model should be that a force F moves an infinitesimal distance $dy$ and then jumps back down that same $dy$ to do it again, staying ever tangential. Note that $dy$ comes from both the translation and the rotation so that the force is doing the two seemingly contradictory things we need it to do: 1. traveling with the surface so it can do all the needed work (enough to provide the energy for rotation and translation), and 2. remaining vertical. Calculations that only move the force the distance that the ball moves will necessarily violate conservation of energy.
Analysis
The torque is a set function of force, $Fr$, so the acceleration and angular acceleration are both proportional to force, the whole way. This constant ratio of $a$ and $\alpha$ means the angular and translational velocities will also be in that same proportion. Or, in math:
$$T=Fr=I \alpha,, F= \tfrac{I \alpha}{r}=ma$$ $$a = \frac{I}{mr} \alpha$$ $$v_0 , \omega_0 =0, a=k \alpha \implies v=k \omega, \forall t$$
(Because $v = \int adt=\int k \alpha dt = k \omega$ without integration constant. At the end of any duration $v_f=k \omega_f$ - meaning $\forall t$)
Kinetic energy:
$$E = \tfrac{1}{2} (mv^2+I \omega^2)= \tfrac{1}{2} [m( \frac{I}{mr} \omega)^2+I \omega^2]$$
$$= \tfrac{1}{2} (\frac{I^2}{mr^2} + I) \omega^2)$$
For the same $ \omega$’s, kinetic energy would be higher for case B, so $I_A <I_B,E_A=E_B \implies \omega_A > \omega_B$. This is not surprising. It says the ball with lower moment of inertia ends up spinning faster. Specifically $$ \frac{\omega_A}{\omega_B} = \sqrt{ \frac{ \tfrac{I_B^2}{m_Br_B^2} + I_B}{\tfrac{I_A^2}{m_Ar_A^2} + I_A}} = \sqrt{ \frac{I_B^2 + I_B m r^2}{I_A^2 + I_A m r^2}}
$$ When $m_A=m_B,r_A=r_B, I_A<I_B$.
Subbing in $ \omega = \frac{mr}{I}v$ instead shows $v_A < v_B$. And we have a similar radical and ratio, except this time based on $m_i+ \tfrac{m_i^2r_i^2}{I_i}$.
Notes
There are several interesting things in comparing $m_i+\tfrac{m_i^2r_i^2}{I_i}$(which is how changes affect $v$), with $I_i+\tfrac{I_i^2}{m_ir_i^2}$(how changes affect $\omega$). For example, why would increasing the mass increase $\omega$ for that case? Shouldn’t larger mass mean more energy is going into accelerating the object than before, with less available for rotation?
First, once under the radical, both include two terms rather than one for how it decreases with their respective inertial resistances ($m, I$). That’s because, for example, increasing $m$ (without changing $I$) has a double effect on $v$: one effect is because the portion of energy that goes toward translation now has to accelerate a larger mass (the $m^2$ in one term), and another effect is from the fact that larger $m$ makes less energy go toward translation as rotation has just become relatively easier (the $m$ term).
It is this latter factor, more energy going toward rotation whenever $m$ increases, that makes $\omega$ increase if $m$ goes up (the $m^{-1}$ term). Because $I$ hasn’t changed, the higher $\omega$ means a higher rotational kinetic energy, and because $E$ hasn’t changed, it also means a higher proportion of energy going to rotation. To summarize: Increasing $m$ increases the mass we need to accelerate linearly and decreases the energy going to translation, the double effect is the intuition for $m$ appearing in two terms in the $v$ ratio. The same for $I$ and $\omega$: $I$ appears in two terms in $\omega$ for the two effects, and is in one term for the single effect on $v$.
In some sense $m$ appearing twice in $v$‘s equation and once in $\omega$’s makes it possible for more overall motion to happen when $m$ is decreased, for the same $E$, (ie one will go up more than the other goes down). Similarly for $I$. This is not the case with radius. Remembering that $I$ is not changing, and increasing radius, we should not have a reduction in overall motion. Increases in $r$ make torque easier to create, but don’t increase energy or decrease overall resistance to motion. This means there will be more energy going to angular acceleration, but since $m$ and $I$ have not changed, $v$ drops a corresponding amount. Hence we have $r$ appearing only once in each ratio, as $r^2$.
