Set up the rolling condition. Each small disk rolls without slipping on the inside of the large disk along the circumference. The contact-point arc lengths covered by the two small disks must, together with the initial gap, total the full circumference traversal needed to meet again.

Geometry of meeting. By symmetry of the angular speeds $\omega$ and $2\omega$, the two contact points move with angular rates on the large disk in inverse proportion. Let the two contact points sweep angles ${\varphi }_1$ and ${\varphi }_2$ on the large disk such that ${\varphi }_1+{\varphi }_2=2\pi -2\Delta \theta$ (they meet after closing the gap on the far side).

Rolling without slipping gives, for each small disk, ${\omega }_{\text{small}}\cdot r={\Omega }_{\text{centre}}\cdot (R+r)$, where ${\Omega }_{\text{centre}}$ is the rate at which the small disk's centre revolves around the large disk's centre. Hence the two centres revolve at ${\Omega }_1=\frac{\omega r}{R+r}$ and ${\Omega }_2=\frac{2\omega r}{R+r}$.

Initial separation. With $r=R/50$, the constraint that the two disks were touching initially gives $\Delta \theta =\frac{2r}{R+r}=\frac{2}{51}$ rad.

Time to meet. The relative angular speed of the two centres about the large disk's centre is ${\Omega }_1+{\Omega }_2=\frac{3\omega r}{R+r}=\frac{3\omega }{51}$. They must close the angular gap $2\pi -2\Delta \theta =2\pi -\frac{4}{51}$ (accounting for the finite size of the small disks on both sides).

$\therefore \tau =\frac{2\pi -4/51}{3\omega /51}=\frac{51\left(2\pi -\frac{4}{51}\right)}{3\omega }$, which matches option (C).