Answer
See the detailed answer below.
Work Step by Step
We know that the cyclotron frequency of a charged particle in a magnetic field is given by
$$f=\dfrac{qB}{2\pi m}$$
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,
$$f=\dfrac{qB}{2\pi (1.6605\times 10^{-27})M}\tag 1$$
So we just have to find the atomic mass of each ion and plug the result into (1).
$$\color{blue}{\bf [a]}$$
For $\rm O_2^+$;
$$f_{\rm O_2^+}=\dfrac{qB}{2\pi (1.6605\times 10^{-27})(2M_{\rm O})} $$
Plug the known;
$$f_{\rm O_2^+}=\dfrac{(1.6022\times 10^{-19})(3.0000)}{2\pi (1.6605\times 10^{-27})(2\times 15.995)} $$
$$f_{\rm O_2^+}=\color{red}{\bf 1.4401}\;\rm MHz$$
$$\color{blue}{\bf [b]}$$
For $\rm N_2^+$;
$$f_{\rm N_2^+}=\dfrac{qB}{2\pi (1.6605\times 10^{-27})(2M_{\rm N})} $$
Plug the known;
$$f_{\rm N_2^+}=\dfrac{(1.6022\times 10^{-19})(3.0000)}{2\pi (1.6605\times 10^{-27})(2\times 14.003)} $$
$$f_{\rm N_2^+}=\color{red}{\bf 1.6450}\;\rm MHz$$
$$\color{blue}{\bf [c]}$$
For $\rm CO^+$;
$$f_{\rm CO^+}=\dfrac{qB}{2\pi (1.6605\times 10^{-27})( M_{\rm C}+M_{\rm O})} $$
Plug the known;
$$f_{\rm CO^+}=\dfrac{(1.6022\times 10^{-19})(3.0000)}{2\pi (1.6605\times 10^{-27})( 12.000+15.995)} $$
$$f_{\rm CO^+}=\color{red}{\bf 1.6457}\;\rm MHz$$
