For the reaction $ A\xrightarrow[{}]{{}} $ products; let the order of reaction is n. Then $ r=k{{[A]}^{n}} $
$ =k\times {{(1.2\times {{10}^{-3}})}^{n}} $ ...(i)
At the initial concentration of $ 3.24\times {{10}^{-2}}M, $ the rate is $ \theta $ times.
Hence, $ 9r=k\times {{(3.24\times {{10}^{-2}})}^{n}} $ ...(ii)
On dividing, $ \frac{9r}{r}=\frac{k\times {{(3.24\times {{10}^{-2}})}^{n}}}{k\times {{(1.2\times {{10}^{-3}})}^{n}}} $
$ 9={{(27)}^{n}} $
$ 9={{({{3}^{3}})}^{2/3}} $
$ \therefore $ $ n=\frac{2}{3} $