On applying Law of mass action over slowest step,

$\mathrm{R}=\mathrm{k}\left[{\mathrm{NOBr}}_2\right]\left[\mathrm{NO}\right]$

The slowest step is the rate determining step.

On applying Law of mass action on equilibrium of Step I

$\mathrm{k}=\frac{\left[{\mathrm{NOBr}}_2\right]}{\left[\mathrm{NO}\right]\left[{\mathrm{Br}}_2\right]}$

$\left[{\mathrm{NOBr}}_2\right]=\mathrm{k}\left[\mathrm{NO}\right]\left[{\mathrm{Br}}_2\right]$

Hence, overall rate equation will be

$\mathrm{R}=\mathrm{k}{\left[\mathrm{NO}\right]}^2\left[{\mathrm{Br}}_2\right]$

$=\mathrm{k}\text{'}{\left[\mathrm{NO}\right]}^2\left[{\mathrm{Br}}_2\right]$

The overall order is equal to sum of the powers of the concentration terms in the rate law.

The overall order of the reaction is 3.