Check each molecule against the octet rule (every non-hydrogen atom surrounded by 8 valence electrons, hydrogen by 2):

$\mathrm{C}{\mathrm{O}}_2$: O=C=O, all atoms octet-complete. $✓$

${\mathrm{C}}_2{\mathrm{H}}_4$: H${}_2$C=CH${}_2$, each C has 8 electrons. $✓$

HCl: H-Cl, Cl has 8, H has 2. $✓$

${\mathrm{O}}_3$: ozone is drawn with all octets satisfied via the standard resonance structures. $✓$

$\mathrm{BC}{\mathrm{l}}_3$: B has only 6 electrons (electron deficient). $\times$

NO: odd-electron species (11 valence electrons total), N has 7. $\times$

$\mathrm{N}{\mathrm{O}}_2$: odd-electron species, N has 7. $\times$

${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$: in the conventional structure S has 12 electrons (expanded octet). $\times$

Option (A): CO${}_2$, C${}_2$H${}_4$, HCl (3 molecules satisfy octet). $✓$

Option (B): O${}_3$ and HCl only (2 satisfy). $\times$

Option (C): none satisfy octet. $\times$

Option (D): CO${}_2$, O${}_3$, C${}_2$H${}_4$ (3 satisfy). $✓$

$\therefore$ Correct: (A) and (D).