${X_2}\xrightarrow{{}}2X$; $\Delta H = + a\,\,kJ/mole$ ………. $(ii)$
${Y_2}\xrightarrow{{}}2Y$; $\Delta H = + 0.5\,a\,\,kJ/mole$ ……….$(iii)$
$\frac{1}{2} \times {\text{(ii) }} + \frac{{\text{1}}}{{\text{2}}} \times {\text{(iii) - (i),}}$ gives
$\Delta H = \left( { + \frac{a}{2} + \frac{{0.5}}{2}a - a} \right)\,\,kJ/mole$
$ + \frac{a}{2} + \frac{{0.5a}}{2} - a = - 200$
$a = 800$