Let the order of reaction with respect to $\mathrm{A}$ be $\mathrm{x}$ and with respect to B be $\mathit{y}$.

According to details given ,
$4.2\times {10}^{-2}=\mathrm{k}{\left[0.2\right]}^{\mathrm{x}}{\left[0.3\right]}^{\mathrm{y}}$ .....(1)
$6.0\times {10}^{-3}=\mathrm{k}{\left[0.1\right]}^{\mathrm{x}}{\left[0.1\right]}^{\mathrm{y}}$ .......(2)
$1.68\times {10}^{-1}=\mathrm{k}{\left[0.4\right]}^{\mathrm{x}}{\left[0.3\right]}^{\mathrm{y}}$ .......(3)
$2.40\times {10}^{-2}=\mathrm{k}{\left[0.1\right]}^{\mathrm{x}}{\left[0.4\right]}^{\mathrm{y}}$ ........(4)
Dividing equation (4) by (2), we get
$\frac{2.40\times {10}^{-2}}{6.0\times {10}^{-3}}=\frac{\mathrm{k}{\left[0.1\right]}^{\mathrm{x}}{\left[0.4\right]}^{\mathrm{y}}}{\mathrm{k}{\left[0.1\right]}^{\mathrm{x}}{\left[0.1\right]}^{\mathrm{y}}}$
$4=\frac{{\left[0.4\right]}^{\mathrm{y}}}{{\left[0.1\right]}^{\mathrm{y}}}$
${\left(4\right)}^1={\left(4\right)}^{\mathrm{y}}$
$\mathit{y}=1$
Dividing equation (1) by (3), we get
$\frac{4.2\times {10}^{-2}}{1.68\times {10}^{-1}}=\frac{\mathrm{k}{\left[0.2\right]}^{\mathrm{x}}{\left[0.3\right]}^{\mathrm{y}}}{\mathrm{k}{\left[0.4\right]}^{\mathrm{x}}{\left[0.3\right]}^{\mathrm{y}}}$
$0.25=\frac{{\left[0.1\right]}^{\mathrm{x}}}{{\left[0.2\right]}^{\mathrm{x}}}$
$\left(0.25\right)={\left(0.5\right)}^{\mathrm{x}}$
${\left(0.5\right)}^2={\left(0.5\right)}^{\mathrm{x}}\mathrm{x}=2$
(a) So the rate of reaction with respect to $\mathit{A}$ is 2 and with respect to $\mathit{B}$ is 1 .
(b) Rate law $=\mathit{k}{\left[\mathit{A}\right]}^2\left[\mathit{B}\right]$ Overall order of reaction is 3 .
(c) Rate constant,
$=\frac{6.0\times {10}^{-3}}{{\left(0.1\right)}^2\left(0.1\right)}$
$=6.0{\mathrm{mol}}^{-2}{\mathrm{L}}^2{\min }^{-1}$