$\text{ If }\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{k}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}^{\mathrm{n}}$and rate become 100 times when pH changes form 2 to 1. Hence order of reactions is 

$$\begin{gathered} \mathrm{pH}-2\to 1 \\ \mathrm{rate}-\mathrm{r}\to 100\mathrm{r} \\ \mathrm{We} \mathrm{know} \mathrm{that}, \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=? \\ 2=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-2} \\ \mathrm{Similarly}, \mathrm{when} \mathrm{pH}=1, \mathrm{then} \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-1} \\ \mathrm{substituting} \mathrm{the} \mathrm{above} \mathrm{values} \mathrm{in} \mathrm{rate}\left(\mathrm{r}\right)=\mathrm{k}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}^{\mathrm{n}} \mathrm{equation} \\ \mathrm{we} \mathrm{have}, {\mathrm{r}}_1=\mathrm{k}{\left[{10}^{-2}\right]}^{\mathrm{n}}=\mathrm{r} \\ {\mathrm{r}}_2=\mathrm{k}{\left[{10}^{-1}\right]}^{\mathrm{n}}=100 \mathrm{r} \\ \therefore \frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}=100={\left(\frac{{10}^{-1}}{{10}^{-2}}\right)}^{\mathrm{n}} \\ {10}^2={10}^{\mathrm{n}} \\ \therefore \mathrm{n}=2 \end{gathered}$$