$\Rightarrow \Delta H _{ p }^0( AB )=-200\,kJ\,mol\,m ^{-1}$

$\therefore \Delta H _{ R }^0$ for reaction $A _2+ B _2 \rightarrow 2 AB$ is $-400\,kJ\,mol ^{-1}$

Given: Bond Enthalpy of $A _2, B _2$ and $AB$ is $1: 0.5: 1$

Assuming bond enthalpy of $A _2$ be $x\, kJ\, mol { }^{-1}$

$\therefore$ Bond enthalpy $B _2=0.5 \times kJ\, mol ^{-1}$

$\therefore$ Bond enthalpy $AB =( x ) kJ\, mol\, m ^{-1}$

$-400= x +0.5 x -2 x$

$-400=-0.5 x$

$\therefore x =800\,kJ / mol$