Can ionic and atomic radii be compared?
A periodic table coupled with simple electrostatic arguments should suffice to provide at least a rough ordering of ions and atoms. For the ions:


We expect POSITIVE ions to be SMALLER than the corresponding atom because there are fewer electrons than protons => an excess of POSITIVE charge => the electron cloud is more attracted to the nucleus on average and becomes more 'concentrated' => $\ce{Na+}$ is 'small' and $\ce{Mg^{2+}}$ is 'very small'.
We expect NEGATIVE ions to be LARGER than the corresponding atom because there are more electrons than protons => an excess of NEGATIVE charge => the electron cloud becomes more diffuse to compensate, lowering the Coulombic interactions => $\ce{F-}$ is 'large', $\ce{O^{2-}}$ is 'very large'.


Even though $\ce{Na+}$ has a lower principal quantum number for its outer shell than $\ce{Mg^{2+}}$, Mg is twice as charged and I believe this can be more important. $\ce{F-}$ and $\ce{O^{2-}}$ have the same principal quantum number for their outer shells, but $\ce{O^{2-}}$ is more charged and accordingly, we expect that $\ce{O^{2-}}$ has a larger radius.

$\ce{Ne}$ carries no unbalanced charge and has comparable valence principal quantum number to the ions, so we expect its radius to fall somewhere between the positive ions (which are smaller) and the negative ions (which are larger). Go over these arguments in your head and try to put the pieces together yourself from here.

(Note: I wouldn't be willing to stake my life on this, so some community validation or corrections would be much appreciated)