Calculating bond energy given the enthalpy value of burning reaction
How can I calculate the enthalpy for those balanced reaction given the data of $\ce{NO +O2 -&gt; NO2}$ enthalpy?


The way the enthalpy data is presented is sloppy (is this the original text of the exercise?). For some types of enthalpy, it is unnecessary to specify the chemical reaction equation because there is a convention. For example, for the enthalpy of formation, it is understood that the substance in question is the only product, the stoichiometry coefficient is one, and that the reactants are elements with coefficients so that the equation is balanced. I am not aware of any such convention for the enthalpy of combustion.

What is the enthalpy of $\ce{2NO(g) + O2(g) -&gt; 2NO2(g)}$?

We can look up the enthalpies of formation for the nitrogen-containining compounds:

$$\Delta H_f(\ce{NO}) = \pu {90.25 kJ/mol}$$
$$\Delta H_f(\ce{NO2}) = \pu {31.18 kJ/mol}$$

From this, we can calculate the reaction enthalpy:

$$\Delta H_r = 2 \cdot \pu {31.18 kJ/mol} - 2 \cdot \pu {90.25 kJ/mol} =  \pu {-118.14 kJ/mol}$$

Which combustion reaction has an enthalpy of -57 kJ/mol?

If we divide the balanced chemical equation by 2, we get:

$$\ce{NO + 1/2 O2 -&gt; NO2}$$

The reaction enthalpy is one half of that of the original reaction, -59.07 kJ/mol. So for the reaction describing the combustion of one mole of NO, the reaction enthalpy is about -57 kJ/mol.

What about the bond dissociation energy?

Everything else in the OP's calculation is correct. If you plug in the correct combustion enthalpy, the calculated bond energy will be correct.