Find oxidation state of nitrogen in each compound using oxygen as $-2$.

For $NO$: $x + (-2) = 0$ $x = +2$ So, oxidation state of $N$ in $NO$ is $+2$.

For $N_2O$: Let average oxidation state of each nitrogen be $x$. $2x + (-2) = 0$ $2x = +2$ $x = +1$ So, average oxidation state of $N$ in $N_2O$ is $+1$.

For $NO_2$: $x + 2(-2) = 0$ $x - 4 = 0$ $x = +4$ So, oxidation state of $N$ in $NO_2$ is $+4$.

For $N_2O_3$: Let average oxidation state of each nitrogen be $x$. $2x + 3(-2) = 0$ $2x - 6 = 0$ $2x = +6$ $x = +3$ So, average oxidation state of $N$ in $N_2O_3$ is $+3$.

Arrange in increasing order: $N_2O(+1) < NO(+2) < N_2O_3(+3) < NO_2(+4)$

Match with options: This corresponds to Option C.

Comparison with stored answer: Stored correct answer = C. My derived answer = C. Hence, they agree.