A heavy ring of mass m is secured on the periphery of a light...

The system consists of a heavy ring of mass $m$ and a small particle of mass $m$.

Since the ring is secured on the periphery of the circular disc, its centre of mass lies at the geometric centre of the disc. The small particle is also secured at the centre of the disc. Therefore, the centre of mass of the entire system coincides with the centre of the disc.

The total mass of the system is:
$M = m_{\text{ring}} + m_{\text{particle}} = m + m = 2m$

The centre of the disc (which is the centre of mass of the system) moves in a circle of radius $r$ with a uniform speed $v$. The centripetal acceleration of the centre of mass is:
$a_{\text{cm}} = \dfrac{v^2}{r}$

According to Newton's second law for a system of particles, the net external force required to maintain this circular motion is given by the product of the total mass and the acceleration of the centre of mass:
$F_{\text{ext}} = M a_{\text{cm}} = 2m \left(\dfrac{v^2}{r}\right) = \dfrac{2mv^2}{r}$

This total external force must act on the system as a whole. Without further information about how the force is applied (e.g., via a string attached to the centre or the edge), we can only conclude that a net external force of $\dfrac{2mv^2}{r}$ must act on the system.