Answer
$-233\text{ kJ}$
Work Step by Step
We combine the following reactions:
$NO_{(g)}+O_{3(g)}\Rightarrow NO_{2(g)}+O_{2(g)}\text{ -199 kJ}$
$O_{(g)}\Rightarrow \frac{1}{2}O_{2(g)}\text{ -247.5 kJ}$
$\frac{3}{2}O_{2(g)}\Rightarrow O_{3(g)}\text{ 213.5 kJ}$
Therefore, the enthalpy of the reaction is:
$\Delta H = -199 -247.5+213.5=-233\text{ kJ}$