The increasing order of the ionic radii of the given isoelectronic species is :

All the given particles are isoelectronic; each one contains $$18$$ electrons (the same as argon).

For an isoelectronic series, the size primarily depends on the effective nuclear charge $$Z_{\text{eff}}$$ experienced by the common electron cloud.
• Larger nuclear charge (more protons) pulls the electrons inward more strongly and gives a smaller ionic radius.
• Smaller nuclear charge exerts a weaker pull, so the ionic radius is larger.

Write the actual nuclear charge (number of protons) for every species:
$$Ca^{2+}: Z = 20$$
$$K^{+}: \;\;Z = 19$$
$$Cl^{-}: Z = 17$$
$$S^{2-}: Z = 16$$

Because the electron count is the same in all four cases, the ionic radius increases as the nuclear charge decreases:

$$Ca^{2+} \;(\!Z=20\!) \lt K^{+} \;(\!Z=19\!) \lt Cl^{-} \;(\!Z=17\!) \lt S^{2-} \;(\!Z=16\!)$$

Hence the correct increasing order of ionic radii is:

$$Ca^{2+},\; K^{+},\; Cl^{-},\; S^{2-}$$

So, the right choice is
Option C which is: $$Ca^{2+}, K^{+}, Cl^{-}, S^{2-}$$.