The idea here is that you can treat the heat needed to get this reaction going as a reactant.
The problem provides you with the following thermochemical equation
#"N"_ (2(g)) + "O"_ (2(g)) + "180 kJ" -> 2"NO"_ ((g))#
Now, notice that
Since the oxygen gas is in excess, the amount of nitrogen as that you have in your sample will determine how much heat should be absorbed in order to get the reaction going and how much nitric oxide will be produced.
Use the molar mass of nitrogen gas to convert the sample to moles
#13.7 color(red)(cancel(color(black)("g"))) * "1 mole N"_2/(28.0134color(red)(cancel(color(black)("g")))) = "0.48905 moles N"_2#
You can now use the fact that every
This will give you
#0.48905 color(red)(cancel(color(black)("moles N"_2))) * "180 kJ"/(1color(red)(cancel(color(black)("mole N"_2)))) = color(darkgreen)(ul(color(black)("88 kJ")))#
I'll leave the answer rounded to two sig figs, the number of sig figs you have for the amount of heat given by the thermochemical equation.