Answer
See the detailed answer below.
Work Step by Step
The particle (ion) undergoes two stages, accelerating in a straight line by the electric field and then moving in a semicircular path at constant speed by the magnetic field.
In the second stage, the ion undergoes a cyclotron motion that has a radius that is given by
$$r=\dfrac{mv}{qB}=\dfrac{mv}{eB}$$
So the speed of the ion is given by
$$v=\dfrac{erB}{m} $$
where $r=D/2$ where $D$ is the diameter needed which is 8 cm, so
$$v=\dfrac{eDB}{2m} \tag 1$$
In the first stage,
Now we need to find the voltage needed to give the particle this speed so that we can use energy conservation.
$$E_i=E_f$$
$$K_i+U_{ie}=K_f+U_{fe}$$
$$0+e\Delta V=\frac{1}{2}m v^2+0$$
Hence,
$$\Delta V=\dfrac{m v^2}{2e} $$
Plug $v$ from (1),
$$\Delta V=\dfrac{m }{2e}\left( \dfrac{eDB}{2m} \right)^2 $$
$$\Delta V= \dfrac{eD^2B^2}{8m} $$
where $m$ is the mass of the element which is given by
$$m=M\times 1.6605\times 10^{-27}$$
where $M$ is the atomic mass of the element. Hence,
$$\Delta V= \dfrac{eD^2B^2}{8M (1.6605\times 10^{-27})} $$
$$\color{blue}{\bf [a]}$$
For $\rm O_2^+$;
$$\Delta V_{\rm O_2^+}= \dfrac{eD^2B^2}{8M_{\rm O_2^+} (1.6605\times 10^{-27})} $$
$$\Delta V_{\rm O_2^+}= \dfrac{eD^2B^2}{8(2M_{\rm O}) (1.6605\times 10^{-27})} $$
Plug the known;
$$\Delta V_{\rm O_2^+}= \dfrac{(1.6022\times 10^{-19})(8.0000\times 10^{-2})^2(200.00\times 10^{-3})^2}{8(2\times 15.995) (1.6605\times 10^{-27})} $$
$$\Delta V_{\rm O_2^+}=\color{red}{\bf 96.519}\;\rm V$$
$$\color{blue}{\bf [b]}$$
For $\rm N_2^+$;
$$\Delta V_{\rm N_2^+}= \dfrac{eD^2B^2}{8M_{\rm N_2^+} (1.6605\times 10^{-27})} $$
$$\Delta V_{\rm N_2^+}= \dfrac{eD^2B^2}{8(2M_{\rm N}) (1.6605\times 10^{-27})} $$
Plug the known;
$$\Delta V_{\rm N_2^+}= \dfrac{(1.6022\times 10^{-19})(8.0000\times 10^{-2})^2(200.00\times 10^{-3})^2}{8(2\times 14.003) (1.6605\times 10^{-27})} $$
$$\Delta V_{\rm N_2^+}=\color{red}{\bf 110.25}\;\rm V$$
$$\color{blue}{\bf [c]}$$
For $\rm CO^+$;
$$\Delta V_{\rm CO^+}= \dfrac{eD^2B^2}{8(M_{\rm C } +M_{\rm O } ) (1.6605\times 10^{-27})} $$
Plug the known;
$$\Delta V_{\rm CO^+}= \dfrac{(1.6022\times 10^{-19})(8.0000\times 10^{-2})^2(200.00\times 10^{-3})^2}{8( 12.000+15.995) (1.6605\times 10^{-27})} $$
$$\Delta V_{\rm CO^+}=\color{red}{\bf 110.29}\;\rm V$$
