Let the rate equation is

$$\text { Rate }=k[A] \times[B]^y$$

Therefore, we can write

$$\begin{aligned} & 2 \times 10^{-3}=k[0.1]^x[0.1]^y \quad \text{..... (i)}\\ & 4 \times 10^{-3}=k[0.2]^x[0.1]^y \quad \text{..... (ii)}\\ & 1.6 \times 10^{-2}=k[0.2]^x[0.2]^y \quad \text{..... (iii)} \end{aligned}$$

$$\begin{aligned} & \text { (ii) } \div \text { (i); } \\ & \frac{4 \times 10^{-3}}{2 \times 10^{-3}}=\frac{k[0.2]^x[0.1]^y}{k[0.1]^x[0.1]^y} \\ & \Rightarrow \quad \frac{2}{1}=\frac{(0.2)^x}{(0.1)^x}=\left(\frac{2}{1}\right)^x \\ & \therefore \quad x=1 \end{aligned}$$

$$\begin{aligned} & \text { (ii) } \div \text { (iii); } \\ & \frac{4 \times 10^{-3}}{1.6 \times 10^{-2}}=\frac{k[0.2]^x[0.1]^y}{k[0.2]^x[0.2]^y} \\ & \Rightarrow \quad \frac{1}{4}=\frac{(0.1)^y}{(0.2)^y}=\left(\frac{1}{2}\right)^y \\ & \therefore \quad y=2 \\ & \therefore \quad \text { Rate }=k[A]^1[B]^2 \end{aligned}$$

First order with respect to A while second order with respect to B.