For electric field,Refer image 1.\\\\(q\\rightarrow\\\\) change of electron.Four due to electric field \\\\(=qE\\\\)\\\\(\ herefore \\dfrac{mv_{0^2}}{R_e}=qE\\\\)where \\\\(R_e\\\\) is radius of curvature,\\\\(\ herefore \\dfrac{1}{R_e}=\\dfrac{qE}{mv_{0^2}}\\\\)\\\\(\ herefore =\\dfrac{mv_{0^2}}{qE}=r_1\\\\)In magnetic field,Focus on electron \\\\(=qV_0B\\\\)Refer image 2.Balancing of centripetal and centrifugal focuses,\\\\(\\dfrac{mV_{0^2}}{Rb}=qV_0B\\\\) \\\\(\ herefore R_b=\\dfrac{mV_0}{qB}=r_2\\\\)\\\\(\ herefore \\dfrac{r_1}{r_2}=\\dfrac{R_c}{R_b}=\\dfrac{\\dfrac{mV_{0^2}}{qE}}{\\dfrac{mV_0}{qB}}\\\\)\\\\(\ herefore \\dfrac{r_1}{r_2}=\\dfrac{BV_0}{E}\\\\)Answer is option \\\\(D\\\\)