1. The electronic configuration

$1s^2 2s^2 2p^6 3s^2 3p^6$

contains $18$ electrons, which is the configuration of argon.

So we need to find which aqueous salt will not contain any ion having 18 electrons.

2. Check the ions formed by each salt in aqueous solution.

Option A: $NaF$

It dissociates as:

$NaF \rightarrow Na^+ + F^-$

• $Na$ has atomic number $11$, so $Na^+$ has $10$ electrons: $1s^2 2s^2 2p^6$
• $F$ has atomic number $9$, so $F^-$ has $10$ electrons: $1s^2 2s^2 2p^6$

Neither ion has $18$ electrons.

Option B: $NaCl$

It dissociates as:

$NaCl \rightarrow Na^+ + Cl^-$

• $Na^+$ has $10$ electrons.
• $Cl$ has atomic number $17$, so $Cl^-$ has $18$ electrons: $1s^2 2s^2 2p^6 3s^2 3p^6$

So this solution does contain an ion with the given configuration.

Option C: $KBr$

It dissociates as:

$KBr \rightarrow K^+ + Br^-$

• $K$ has atomic number $19$, so $K^+$ has $18$ electrons.
• $Br^-$ has $36$ electrons.

So this solution does contain an ion with the given configuration.

Option D: $CaI_2$

It dissociates as:

$CaI_2 \rightarrow Ca^{2+} + 2I^-$

• $Ca$ has atomic number $20$, so $Ca^{2+}$ has $18$ electrons.
• $I^-$ has $54$ electrons.

So this solution does contain an ion with the given configuration.

3. Therefore, the only salt whose aqueous solution does not contain any ion with argon configuration is:

$\boxed{NaF}$

So, the correct option is A.