For a chemical reaction, A ~ B, the rate of the reaction is $2\times {10}^{-3}\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{s}}^{-}$ when the initial concentration is $0.25\mathrm{mol}{\mathrm{dm}}^{-3}$ The rate of the same reaction is $1.6\times {10}^{-2}{\mathrm{moldm}}^{-3}{\mathrm{s}}^{-}$ when the initial concentration is0.1 $\text{ mol }{\mathrm{dm}}^{-3}$.The order of the reaction is:

We know that,rate = k [A]" where R is rate constant is order of reaction and [A] is concentration of reactant.

Given:Rate, $=2\times {10}^{-3}\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{s}}^{-};\left[A_0\right]=0.05\mathrm{mol}{\mathrm{dm}}^{-3}$

$\begin{array}{r}{\text{ Rate }}_{\text{II }}=1.6\times {10}^{-2}\mathrm{mol}{\mathrm{dm}}^{-3}{\mathrm{s}}^{-};\left[A_0\right]=0.1\mathrm{mol}{\mathrm{dm}}^{-3}\\ \therefore 2\times {10}^{-3}=k[0.05{]}^n-(1);1.6\times {10}^{-2}=k[0.1{]}^n\end{array}$

Dividing (1) by (2),we get; $\frac{2\times {10}^{-3}}{1.6\times {10}^{-2}}={\left(\frac{0.05}{0.10}\right)}^n$

or $\frac{(0.05{)}^n}{(0.05{)}^n\times 2^n}\text{ or }\frac{1}{2^n}=\frac{1}{8}\text{ or }\frac{1}{2^3}.\text{ Hence, }n=3$

So,the correct answer is (b)