Two particles X and Y having equal charges are being accelerated...

Energy gained due to potential difference by charged particle:

$KE = qV$

Relation between momentum & kinetic energy:

$p = \sqrt{2m\left(KE\right)}$

Now, radius of the circular path under magnetic field will be given by, 

$R = \dfrac{mv}{qB} = \dfrac{p}{qB} = \dfrac{\sqrt{2m ( KE )}}{qB} = \dfrac{\sqrt{2mqV}}{qB}$

Therefore,

$R ∝ \sqrt{m} ⇒ m ∝ {R}^{2}$

$⇒ \dfrac{{m}_{1}}{{m}_{2}} = {\left(\dfrac{{R}_{1}}{{R}_{2}}\right)}^{2}$