Two particles X and Y carrying the same charge are accelerated...

The kinetic energy gained by a particle of charge $q$ accelerated through a potential difference $V$ is $K = qV$.

The momentum of the particle is $p = \sqrt{2mK} = \sqrt{2mqV}$.

When the particle enters a uniform magnetic field $B$ perpendicularly, the radius of its circular path is given by $R = \dfrac{p}{qB} = \dfrac{\sqrt{2mqV}}{qB}$.

Squaring both sides, we get $R^2 = \dfrac{2mV}{qB^2}$, which gives $m = \dfrac{qB^2 R^2}{2V}$.

Since $q$, $V$, and $B$ are the same for both particles $X$ and $Y$, the mass is directly proportional to the square of the radius, $m \propto R^2$.

Therefore, the ratio of their masses is $\dfrac{m_X}{m_Y} = \left(\dfrac{R_1}{R_2}\right)^2$.