Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore, rate of the reaction is given by,  

$Rate =k{\left[A\right]}^x{\left[B\right]}^y$

$6.\times {10}^{-3}=k{\left[0.1\right]}^x{\left[0.1\right]}^y \left(i\right)$
$7.2\times {10}^{-2}=k{\left[0.3\right]}^x{\left[0.2\right]}^y \left(ii\right)$
$2.88\times {10}^{-1}=k{\left[0.3\right]}^x{\left[0.4\right]}^y \left(iii\right)$
$2.40\times {10}^{-2}=k{\left[0.4\right]}^x{\left[0.1\right]}^y \left(iv\right)$

Dividing equation (iv) by (i), we obtain

$\frac{2.40\times 10}{6.{010}^{-3}}=\frac{K{\left[0.4\right]}^x{\left[0.1\right]}^y}{K{\left[0.1\right]}^x{\left[0.1\right]}^y}$
$\Rightarrow 4=\frac{{\left[0.4\right]}^x}{{\left[0.1\right]}^x}$
$\Rightarrow 4={\left(\frac{0.4}{0.1}\right)}^x$
$\Rightarrow {\left(4\right)}^1=4^x$
$\Rightarrow x=1$
$$

Dividing equation (iii) by (ii), we obtain

$\frac{2.88\times {10}^{-1}}{7.2\times {10}^2}\frac{K{\left[0.3\right]}^x{\left[0.4\right]}^y}{K{\left[0.3\right]}^x{\left[0.2\right]}^y}$
$\Rightarrow 4={\left(\frac{0.4}{0.2}\right)}^y$
$\Rightarrow 4=2^y$
$\Rightarrow 2^2=2^y$
$\Rightarrow y=2$

Therefore, the rate law is
Rate = k [A] [B]${}^2$

$K=\frac{Rate}{\left[A\right]{\left[B\right]}^2}$

From experimental I, we obtain

$$
$K=\frac{6.0\times L^{2}mo{l}^{-2}mi{n}^{-1}}{(0.1 mol L^2){(0.1 mol L^{-1})}^2}$
$=6.0 L^3 mo{l}^{-2} mi{n}^{-1}$

From experimental II, we obtain

$K=\frac{7.2\times {10}^{-2} mol L^{-1} mi{n}^{-1}}{(0.3 mol L^{-1})(0.2 mo{l}^{-1} mi{n}^{-1})}$2
$=6.0 L^2 mo{l}^{-2} mi{n}^{-1}$

From experimental III, we obtain

$k=\frac{2.88\times {10}^{-1} mol L^{-1} mi{n}^{-1}}{(0.3 mol L^{-1} (0.4 mol L^{-1})}$
$=6.0 L^2 mo{l}^{-2} mi{n}^{-1}$
$$

From experimental IV, we obtain

$\mathrm{k} = \frac{2.40\times {10}^{-2} {\mathrm{mol}}^{-1} {\min }^{-1} }{(0.4 \mathrm{mol} {\mathrm{L}}^{-1})}$

=6.0 L2 mol-2 min-1

Therefore, rate constant, k = $6.0 L^2 mo{l}^{-1} mi{n}^{-1}$