The correct increasing order of the ionic radii is:

🧠 All four ions carry 18 electrons — only nuclear charge changes $\ce{Ca^{2+}}$ ($Z=20$), $\ce{K^+}$ ($Z=19$), $\ce{Cl^-}$ ($Z=17$), $\ce{S^{2-}}$ ($Z=16$). Same 18-electron cloud in every case. With more protons pulling on the same electrons, the cloud gets squeezed inward. So size depends on Z alone — higher Z means smaller ion.

🗺️ Rank by decreasing Z $\ce{Ca^{2+}}$ has the most protons (20) → strongest pull → smallest. Then $\ce{K^+}$ (19). Then $\ce{Cl^-}$ (17). Then $\ce{S^{2-}}$ (16) → weakest pull → largest.

Increasing size: $\ce{Ca^{2+}} < \ce{K^+} < \ce{Cl^-} < \ce{S^{2-}}$.

⚠️ The trap Many students see $\ce{K^+}$ and $\ce{S^{2-}}$ and apply the periodic trend — "S is below Cl, so $\ce{S^{2-}}$ should be bigger than $\ce{Cl^-}$" — and end up at option (b). The trick is to first check the electron count. Once all four ions are isoelectronic, forget the period and group trend. Just count protons and rank by Z.

$\boxed{\text{Answer: (d)}}$