$A+B \longrightarrow \text { Products }$

On doubling the initial concentration of $A$ only, the rate of the reaction is also doubled, therefore

$\text {Rate } \propto[A]^{1}$ ..... $(i)$

Let initial rate law is

$\text {Rate}=k[A][B]^{y}$ ..... $(ii)$

If concentration of $A$ and $B$ both are doubled, the rate gets changed by a factor of $8$

$8 \times \text {rate}=k[2 A][2 B]^{y}$ ..... $(iii)$

$\left[\therefore \text { Rate } \propto[A]^{1}\right]$

Dividing Eq. $(iii)$ by Eq. $(ii)$, we get

${8=2 \times 2^{y}} $

${4=2^{y}} $

${(2)^{2}=(2)^{y}} $

${y=2}$

Hence, rate law is, rate $=k[A][B]^{2}$