Let say, rate of reaction = $\mathit{K}{\left[\mathit{A}\right]}^{\mathit{x}}{\left[\mathit{B}\right]}^{\mathit{y}}$.
then,
$1.89\times {10}^{-4}$ = $\mathit{K}{\left[3.0\times {10}^{-2}\right]}^{\mathit{x}}{\left[2\times {10}^{-2}\right]}^{\mathit{y}}$ _____(i)
$1.89\times {10}^{-4}=\mathit{K}{\left[3\times {10}^{-2}\right]}^{\mathit{x}}{\left[4\times {10}^{-2}\right]}^{\mathit{y}}$ _____(ii)
and $7.56\times {10}^{-4}=\mathit{K}{\left[6\times {10}^{-2}\right]}^{\mathit{x}}{\left[4\times {10}^{-2}\right]}^{\mathit{y}}$ _____(iii)
from equation (i) and (ii) , $\left[\mathit{i}\mathit{i}÷\mathit{i}\right]$ we got
$1=2^{\mathit{y}}$
$\Rightarrow 2\circ =2^{\mathit{y}}$
$\Rightarrow \mathit{y}=0$
from equation $\left[\mathit{i}\mathit{i}\mathit{i}÷\mathit{i}\mathit{i}\right]$, we got
$4=2^{\mathit{x}}$
$2^2=2^{\mathit{x}}$
$\mathit{x}=2$
therefore, the order of reaction with respect to A = 2 and w.r.to B = O