The kinetic data for the reaction: $2A+B_2\to 2AB$ are as given below:

Hence the order of reaction w.r.t. A and $B_2$ are, respectively,

$2.5\mathrm{x}{10}^{-3}=\mathrm{K}{\left[2\right]}^{\mathrm{a}}{\left[1.0\right]}^{\mathrm{b}} \dots \left(1\right)$

$1.25 \mathrm{x} {10}^{-3} = \mathrm{K}{[1.0]}^{\mathrm{a}} {[1.0]}^{\mathrm{b}} \dots \left(2\right)$

$1\times {10}^{-2} = \mathrm{K}{\left[2\right]}^{\mathrm{a}} {[2.0]}^{\mathrm{b}} \dots \left(3\right)$

Dividing equation (1) and (2)

$$\begin{gathered} \frac{2.5\times {10}^{-3}}{1.25\times {10}^{-3}}={\left[\frac{2.0}{1.0}\right]}^{\mathrm{a}} \\ \frac{2.0}{1.0}={\left[\frac{2.0}{1.0}\right]}^{\mathrm{a}} \\ \mathrm{a}=1 \end{gathered}$$

Dividing equation (1) and (3)

$\frac{2.5\times {10}^{-3}}{1\times {10}^{-2}}={\left(\frac{1.0}{2.0}\right)}^b$

$$\begin{gathered} \frac{1}{4}={\left(\frac{1}{2}\right)}^{\mathrm{b}} \\ \mathrm{b}=2 \end{gathered}$$