The option(s) in which at least three molecules follow Octet Rule is(are)

For a molecule to obey the octet rule, every atom (especially the central atom) must possess exactly $$8$$ valence electrons (4 electron pairs) after the Lewis structure is drawn.
Two types of violation are common:

(i) Incomplete octet - the central atom has <8 electrons, e.g. $$BF_3$$, $$BCl_3$$.
(ii) Expanded octet - the central atom has >8 electrons, e.g. $$PF_5$$, $$H_2SO_4$$.
(iii) Odd-electron (free-radical) structures automatically break the octet rule because one atom necessarily has 7 electrons; e.g. $$NO$$, $$NO_2$$.

We now examine the four options, molecule by molecule.

• $$CO_2$$ - each O has 8 electrons, the central C has 8 electrons ⇒ obeys.
• $$C_2H_4$$ (ethene) - both C atoms achieve an octet through a double bond and two C-H σ-bonds ⇒ obeys.
• $$NO$$ - 11 valence electrons, one electron remains unpaired; N has only 7 electrons ⇒ violates.
• $$HCl$$ - Cl has 8 electrons, H needs only 2 ⇒ obeys.

Total obeying = 3 out of 4 ⇒ Option A satisfies the condition.

• $$NO_2$$ - odd number (17) of valence electrons; one O has 7 electrons ⇒ violates.
• $$O_3$$ - 18 electrons; every O ends with 8 electrons ⇒ obeys.
• $$HCl$$ - obeys (see above).
• $$H_2SO_4$$ - S forms six σ-bonds (4 S-O + 2 S-O(H)); S gets 12 electrons ⇒ expanded octet ⇒ violates.

Total obeying = 2 ⇒ Option B does not qualify.

• $$BCl_3$$ - B ends with only 6 electrons ⇒ violates.
• $$NO$$ - violates (see above).
• $$NO_2$$ - violates (see above).
• $$H_2SO_4$$ - violates (expanded octet).

Total obeying = 0 ⇒ Option C does not qualify.

• $$CO_2$$ - obeys (see Case A).
• $$BCl_3$$ - B has 6 electrons ⇒ violates.
• $$O_3$$ - obeys (see Case B).
• $$C_2H_4$$ - obeys (see Case A).

Total obeying = 3 ⇒ Option D satisfies the condition.

Hence, the options containing at least three molecules that follow the octet rule are:
Option A and Option D.

Final Answer: Option A (CO₂, C₂H₄, NO, HCl) and Option D (CO₂, BCl₃, O₃, C₂H₄).