The enthalpy of formation of $N_{2}O\left(g\right)$ and NO(g) are 82 and 90 KJ/mol respectively. The enthalpy of the reaction $2N_{2}O\left(g\right)+O_2\left(g\right) \to 4NO\left(g\right)$  is

The reaction of formation of $N_{2}O\left(g\right)$is given below:

$N_2+\frac{1}{2}{O}_2\to N_{2}O...........\left(1\right)$ 

The enthalpy of this reaction is 82 KJ/mol.

The reaction of formation of NO(g) is given below:

$\frac{1}{2}{N}_2+\frac{1}{2}{O}_2\to NO...........\left(2\right)$ 

The enthalpy of this reaction is 90 KJ/mol.

Multiplying the equation (1) by 2 and equation (2) by 4, we get:

$2N_2+O_2\to 2N_{2}O...........\left(3\right)$

The enthalpy of this reaction is 164 KJ/mol.

$2N_2+2O_2\to 4NO...........\left(4\right)$ 

The enthalpy of this reaction is 360 KJ/mol.

 Subtracting equation (3) from (4), we get the required reaction in the question. So, the enthalpy of the reaction can be calculated as:

$\Delta H=360-164=196\text{ KJ}$ 

Therefore, the correct option is D.