Answer
$S^{2-}$ is larger than $K^+$ since the electrons in $S^{2-}$ experience less attraction to the nucleus than those in $K^+$, giving them more freedom and more chance to be further from the nucleus.
Work Step by Step
Both $S^{2-}$ and $K^+$ have 18 electrons, which means the electron-electron repulsions, or in other words, the screening constant in both ions are quite the same.
However, the nuclear charge of $K^+$ is greater than that of $S^{2-}$. That means the attraction force of the $K^+$ nucleus to the electrons is greater than that of the $S^{2-}$ nucleus.
Therefore, though having the same number of electrons, the electrons in $K^+$ are more attracted and as a result, closer to the nucleus than those in $S^{2-}$.
That means the ionic radium of $K^+$ is smaller than the ionic radium of $S^{2-}$. In other words, $S^{2-}$ is larger than $K^+$.