1. We need the order of bond dissociation energy of the $E-H$ bond in group 16 hydrides:

$\mathrm{H_2O,\\ H_2S,\\ H_2Se,\\ H_2Te}$

where:

• $A = O$
• $B = S$
• $C = Se$
• $D = Te$

2. Down group 16, the size of the central atom $E$ increases:

$O < S < Se < Te$

3. As atomic size increases, the overlap between the orbitals of $E$ and $H$ becomes poorer.

• $O-H$ bond has the best overlap and is shortest, so it is strongest.
• Moving down the group, $S-H$, $Se-H$, and $Te-H$ bonds become progressively longer and weaker.

Therefore, bond dissociation energy decreases down the group:

$O-H > S-H > Se-H > Te-H$

4. Translating this into the given labels:

$A > B > C > D$

5. Now check the options:

• Option A: $A > B > C > D$ ✅
• Option B: $D > C > B > A$ ❌
• Option C: $B > A > C > D$ ❌
• Option D: $A > B > D > C$ ❌

Hence, the correct option is A.