Test with freshly prepared ferrous sulphate in acidic medium

Formation of a dark brown ring is the classic brown ring test for the nitrate ion.

$\mathrm{NO_3^-} \xrightarrow[acidic\ medium]{FeSO_4} \text{brown ring complex}$

So, one of the anions present must be:

$\boxed{\mathrm{NO_3^-}}$

Test with neutral $\mathrm{FeCl_3}$

A deep red colour that disappears on boiling with formation of a brown-red precipitate is characteristic of the acetate ion.

• Acetate gives a deep red colour with neutral ferric chloride due to formation of ferric acetate.
• On boiling, ferric acetate hydrolyses, so the red colour disappears and a basic ferric acetate / ferric hydroxide type brown-red precipitate forms.

Thus, the second anion is:

$\boxed{\mathrm{CH_3COO^-}}$

Check the options

• A: $\mathrm{SO_3^{2-}}$ and $\mathrm{C_2O_4^{2-}}$
  No nitrate present, so brown ring test cannot occur. ❌

• B: $\mathrm{CH_3COO^-}$ and $\mathrm{NO_3^-}$
  Nitrate explains brown ring test; acetate explains ferric chloride red colour disappearing on boiling. ✅

• C: $\mathrm{C_2O_4^{2-}}$ and $\mathrm{NO_3^-}$
  Nitrate fits first test, but oxalate does not give the characteristic deep red ferric chloride test described. ❌

• D: $\mathrm{SO_3^{2-}}$ and $\mathrm{CH_3COO^-}$
  No nitrate, so brown ring test cannot occur. ❌

Final answer

The mixture contains:

$\boxed{\mathrm{CH_3COO^-} \text{ and } \mathrm{NO_3^-}}$

Hence, the correct option is:

$\boxed{\text{B}}$