Answer
$rate=k[NO]^2[O_2]$
Work Step by Step
If we compare Experiments 1 and 2, we can see that $[O_2]$ is held constant while $[NO]$ is doubled. And because doubling NO concentration leads to a fourfold increase in the overall rate of the reaction, the rate is proportional to $[NO]^2$ (the rate is in second-order in $NO$).
In contrast, between Experiments 2 and 3, $[NO]$ is kept constant while $[O_2]$ is increased by a factor of 2. Since the reaction rate also increased by a factor of 2, the rate is proportional to $[O_2]$ (the rate is in first-order in $[O_2]$.
Thus, the rate law for this reaction is: $rate=k[NO]^2[O_2]$
