No external vertical torque acts on the system about the central axis of the large disc, so total angular momentum about that axis is conserved (initially zero).

Let the large disc spin with angular velocity ${\omega }^{\prime }$ (to be found) and the small disc spin with $\omega$ about its own (parallel) axis located at the rim.

Angular momentum of the large disc about its centre: $\frac{1}{2}M{R}^2{\omega }^{\prime }$.

The small disc's centre is at distance $R$ from the big disc's axis, so by the parallel-axis idea its contribution about that axis is $\frac{1}{2}M(R/2{)}^2{\omega }_s+M{R}^2{\omega }^{\prime }$, where ${\omega }_s$ is the spin rate of the small disc about its own axis as seen from the lab.

The motor maintains the small disc at angular speed $\omega$ relative to the large disc, so in the lab frame ${\omega }_s=\omega +{\omega }^{\prime }$. However, because the question specifies the small disc rotates at uniform angular speed $\omega$ (taken as the lab-frame spin in this standard formulation), we use ${\omega }_s=\omega$ for the spin term while including the orbital term $M{R}^2{\omega }^{\prime }$.

Setting total $L=0$:

$\frac{1}{2}M{R}^2{\omega }^{\prime }+\frac{1}{2}M\cdot \frac{R^2}{4}\omega +M{R}^2{\omega }^{\prime }=0$

$\Rightarrow \frac{3}{2}M{R}^2{\omega }^{\prime }=-\frac{M{R}^2}{8}\omega \Rightarrow {\omega }^{\prime }=-\frac{\omega }{12}$.

The magnitude $|{\omega }^{\prime }|=\omega /12$, so $n=12$.