The radius of the circular path, of a charges particle in magnetic field
$r=\frac{m v}{q B}=\sqrt{\frac{2 m E}{q B}}$
Here, kinetic energy for proton and helium is same and both are moving in the same magnetic field
$\therefore r \propto \frac{\sqrt{m}}{q}$
So, $\frac{r_{p}}{r_{ e }}=\frac{\frac{\sqrt{m_{p}}}{q_{p}}}{\frac{\sqrt{m_{ He }}}{q_{ He }}}=\sqrt{\frac{m_{p}}{m_{ He }}} \times \frac{q_{ He }}{q_{p}}=\sqrt{\frac{m}{4 m}} \cdot \frac{2 q}{q}=\frac{1}{1}$