The correct increasing order of the ionic radii is

All the given ions are isoelectronic, i.e. each of them possesses $$18$$ electrons:

$$S^{2-}\;(Z = 16,\; 16 + 2 = 18\ \text{electrons})$$
$$Cl^- \;(Z = 17,\; 17 + 1 = 18\ \text{electrons})$$
$$K^+ \;(Z = 19,\; 19 - 1 = 18\ \text{electrons})$$
$$Ca^{2+}\;(Z = 20,\; 20 - 2 = 18\ \text{electrons})$$

For an isoelectronic series the principal quantum number $$n$$ is the same, so the ionic radius depends mainly on the effective nuclear charge $$Z_{\text{eff}}$$ experienced by the electrons.

Rule: Within an isoelectronic series, ionic radius decreases as the atomic number $$Z$$ (and hence $$Z_{\text{eff}}$$) increases, because the nucleus pulls the same number of electrons more strongly.

Ordering the ions by increasing atomic number:

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

Therefore the radii follow the opposite trend:

$$\text{smallest radius} \; Ca^{2+} \lt K^+ \lt Cl^- \lt S^{2-} \; \text{largest radius}$$

Writing the ions in increasing order of ionic radii:

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

This matches Option D.

Final Answer: Option D