#### Answer

The order of the ions in increasing size as follows: $$K^{+}\lt Cl^-\lt S^{2-}$$

#### Work Step by Step

The electron configuration of $K$ is $[Ar]3s^1$. According to octet rule, $K$ will tend to give away 1 electron in $3s$ so that it can reach the stable configuration of the nearest noble gas, which is $Ar$. So the most common ion of $K$ is $K^+$.
Similarly, the electron configuration of $Cl$ is $[Ne]2s^22p^5$. According to octet rule, $Cl$ will tend to receive 1 electron to add to the $2p$ orbital so that it can reach the stable configuration of the nearest noble gas, which is $Ar$, then every possible space for electrons will be filled. So the most common ion of $Cl$ is $Cl^-$.
The electron configuration of $S$ is $[Ne]2s^22p^4$. $S$ will tend to receive 2 electrons to add to the $2p$ orbital so that it can reach the stable configuration of the nearest noble gas, which is $Ar$. So the most common ion of $S$ is $S^{2-}$.
All of the above ions have the same configuration of the noble gas $Ar$. Therefore, they all have 18 electrons. That means the electron-electron repulsions of all three are somewhat the same.
However, the nuclear charge of each of these three is not the same. $K^+$ has the highest nuclear charge, then $Cl^-$ and the lowest is $S^{2-}$. Therefore, the nucleus of $K^+$ attracts electrons more than $Cl^-$, and the nucleus of $Cl^-$ attracts electrons more than $S^{2-}$.
That means we can write the order of the ions in increasing size as follows: $$K^{+}\lt Cl^-\lt S^{2-}$$