1. Test with $BaCl_2$

When $BaCl_2$ is added to the water extract, a white precipitate forms. This suggests formation of an insoluble barium salt.

Let us check the options:

• $I^-$: $BaI_2$ is soluble in water, so no white precipitate.
• $SO_3^{2-}$: forms $BaSO_3$, which is a white precipitate.
• $S^{2-}$: $BaS$ is soluble, so no precipitate.
• $NO_2^-$: barium nitrite is soluble, so no precipitate.

So the likely anion is $SO_3^{2-}$.

2. Reaction of the white precipitate with dilute $HCl$

If the precipitate is $BaSO_3$, then with dilute $HCl$:

$BaSO_3 + 2HCl \rightarrow BaCl_2 + H_2O + SO_2 \uparrow$

The gas evolved is $SO_2$, which has a characteristic pungent/choking odour.

This matches the statement in the question.

3. Rejecting other options

• $I^-$ does not give a white precipitate with $BaCl_2$.
• $S^{2-}$ would give $H_2S$ with acid, but the required white precipitate with $BaCl_2$ is not obtained.
• $NO_2^-$ may give nitrogen oxides with acid, but again no corresponding insoluble white barium precipitate is expected.

4. Conclusion

Therefore, the anion present is:

$\boxed{SO_3^{2-}}$

So, Option B is correct.