Step 1: Understand the goal.
We must find the neutral complex, meaning the whole complex has zero overall charge and needs no extra ions outside it.
Step 2: Look at the obviously charged ones.
Option 3 is $[\text{Cu(NH}_3)_4]^{2+}$, which clearly carries a $+2$ charge. Option 4 is $[\text{Fe(CN)}_6]^{4-}$, which carries a $-4$ charge. So both are charged, not neutral.
Step 3: Look at option 1.
$\text{Na}_3[\text{AlF}_6]$ has three sodium ions outside, which means the complex part $[\text{AlF}_6]^{3-}$ is negatively charged. So this is not neutral.
Step 4: Test option 2 by adding charges.
In $[\text{Co(NO}_2)_3(\text{NH}_3)_3]$, cobalt is $+3$. Each $\text{NO}_2$ is $-1$, and there are three, giving $-3$. Each $\text{NH}_3$ is neutral, giving 0.
Step 5: Total the charges.
\[ (+3) + 3(-1) + 3(0) = 0 \]
The charges cancel, so this complex is neutral with no ions needed outside.
Step 6: Pick the answer.
The neutral complex is $[\text{Co(NO}_2)_3(\text{NH}_3)_3]$, which is option 2.
\[ \boxed{[\text{Co(NO}_2)_3(\text{NH}_3)_3]} \]