Ionic Radii of Isoelectronic Species

Among isoelectronic monoatomic species, size is inversely proportional to atomic number. Hence among isoelectronic species $Na^+$, $O^{2-}$, $N^{3-}$, $F^-$ (having nearest noble gas configuration); Order of size is $Na^+<F^-<O^{2-}<N^{3-}$ $N^{3-}$ has least atomic number, hence largest size.

For isoelectronic species, size $\propto \frac{1}{\text{effective nuclear charge}}$.
Nuclear charge: Ca (20) > K (19) > Ar (18), so radii increase as $Ca^{2+} < K^+ < Ar$.

For isoelectronic species, a greater positive charge means a greater effective nuclear charge and a smaller radius.
So a smaller positive charge on the cation gives a larger, not smaller, ionic radius (e.g. $Na^+ > Mg^{2+} > Al^{3+}$).
The other three statements are correct, so this statement is wrong.

All five species, $Ca^{2+}$, $K^+$, $Ar$, $Cl^-$ and $S^{2-}$, have 18 electrons; they are isoelectronic.
For isoelectronic species the number of electrons is the same, so the size is decided by the nuclear charge: greater the nuclear charge, stronger the pull on the electrons and smaller the size.
Nuclear charges: $Ca^{2+}$ = 20, $K^+$ = 19, $Ar$ = 18, $Cl^-$ = 17, $S^{2-}$ = 16.
So the size of an anion increases as the negative charge increases, whereas the size of a cation decreases with increase in positive charge; the radii of the anions are more than those of the cations.
Thus the correct order is: $Ca^{2+} \lt K^+ \lt Ar \lt Cl^- \lt S^{2-}$

All five ions have 10 electrons (the neon configuration $1s^2\,2s^2\,2p^6$).
Their nuclear charges are: O = 8, F = 9, Na = 11, Mg = 12, Al = 13.
For the same number of electrons, a larger nuclear charge pulls the electron cloud more strongly, so the ion becomes smaller.
Ionic radii (pm): $O^{2-}$ 140, $F^-$ 133, $Na^+$ 102, $Mg^{2+}$ 72, $Al^{3+}$ 54.
So the ionic radii show a significant decrease from $O^{2-}$ to $Al^{3+}$.