#### Answer

An electron in $n=3$ of Kr would experience greater effective nuclear charge than that of Ar and therefore, is closer to the nucleus.

#### Work Step by Step

The nuclear charge of Ar is $Z(Ar)=18+$ and that of Kr is $Z(Kr)=36+$. So $Z(Kr)$ is much greater than $Z(Ar)$.
An electron in a particular shell can only be screened mostly by electrons in the inner shells and a little by electrons in the same shell, and cannot be screened by electrons in the outer shell. Therefore, an electron in $n=3$ are mostly screened by electrons in $n=1$ and $n=2$ and a little screened by electrons in $n=3$.
However, we see that both Kr and Ar have the same number of electrons in $n=1$ (2), $n=2$ (8) and $n=3$ (8). Therefore, $S(Kr)\approx S(Ar)$.
According to the formula, $$Z_{eff}=Z-S$$
we can deduce that $Z_{eff}(Kr)$ is much greater than $Z_{eff}(Ar)$ for an electron in $n=3$.
Therefore, since an electron experiencing greater effective nuclear charge would experience more attraction from the nucleus and stay closer to the nucleus, an electron in $n=3$ of Kr would be closer to the nucleus.