K2Cr2O7 paper acidified with dilute H2SO4 turns green when exposed to:

Concept: Oxidising action of $\text{K}_2\text{Cr}_2\text{O}_7$

Acidified $\text{K}_2\text{Cr}_2\text{O}_7$ (orange) is a strong oxidising agent. When it oxidises a reducing agent, $\text{Cr}^{6+}$ is reduced to $\text{Cr}^{3+}$ (green).

Reaction with $\text{SO}_2$: $\text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 + 3\text{SO}_2 \rightarrow \text{K}_2\text{SO}_4 + \text{Cr}_2(\text{SO}_4)_3 + \text{H}_2\text{O}$

• $\text{Cr}^{6+}$ (orange) → $\text{Cr}^{3+}$ (green) ✓
• $\text{SO}_2$ (+4) → $\text{SO}_4^{2-}$ (+6) — oxidised

Why others don't cause this:

• $\text{CO}_2$: not a reducing agent
• $\text{SO}_3$: not a reducing agent (already in +6 state)
• $\text{H}_2\text{S}$: is a reducing agent but turns the paper yellow/turbid (S deposits), not green

Answer: Option (4) — Sulphur dioxide