$\therefore$ $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{qV}$

or $\quad \mathrm{v}=\sqrt{\frac{2 \mathrm{qV}}{\mathrm{m}}}$

As the charged particle describes a circular path of radius $\mathrm{R}$ in the uniform magnetic field.

$\therefore \quad \frac{\mathrm{mv}^{2}}{\mathrm{R}}=\mathrm{qvB}$

or $\quad \mathrm{R}=\frac{\mathrm{mv}}{\mathrm{qB}}=\frac{\mathrm{m}}{\mathrm{qB}} \sqrt{\frac{2 \mathrm{qV}}{\mathrm{m}}}=\frac{\sqrt{\mathrm{m}}}{\mathrm{B}} \cdot \sqrt{\frac{2 \mathrm{V}}{\mathrm{q}}}$

As $\mathrm{q}, \mathrm{B}$ and $\mathrm{V}$ remain the same.

$\therefore \quad \mathrm{R} \propto \sqrt{\mathrm{m}}$

$\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{A}}}{\mathrm{m}_{\mathrm{B}}}}$

$\Rightarrow \quad \frac{\mathrm{m}_{\mathrm{A}}}{\mathrm{m}_{\mathrm{B}}}=\left(\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}\right)^{2}=\left(\frac{2}{3}\right)^{2}=\frac{4}{9}$