First write balanced reaction for the formation of one mole of AB with the help of diatomic molecule:
21A2+21B2→ABΔHf=−200kJ/mol
Now,
ΔHreaction=21ΔHA−A+21ΔHB−B−ΔHAB(let,ΔHAB=x)
21x+21(0.5x)−x=−200⇒2x+0.25x−x=−200
⇒−0.25x=−200⇒x=800kJ/mol
Bond enthalpy =800kJ/mol.