If X has a poisson distribution such that

$\mathrm{P}\left(X=2\right)=\frac{2}{3}\mathrm{P}\left(X=1\right) \mathrm{then} \mathrm{P}\left(\mathrm{X}=3\right)$ is

Given X follows Poisson distribution 

Given that $\mathrm{P}(X=2)=\frac{2}{3} \mathrm{P}(X=1)$

$$\begin{gathered} \Rightarrow \frac{{\mathrm{e}}^{-\mathrm{\lambda }}.{\mathrm{\lambda }}^2}{2!}=\frac{2}{3} \left(\frac{{\mathrm{e}}^{-\mathrm{\lambda }}.{\mathrm{\lambda }}^1}{1!}\right) \\ \Rightarrow \frac{\mathrm{\lambda }}{2}=\frac{2}{3} \\ \Rightarrow \mathrm{\lambda }=\frac{4}{3} \end{gathered}$$

Now  $\mathrm{P}(X=3)=\frac{{\mathrm{e}}^{-\mathrm{\lambda }}.{\mathrm{\lambda }}^3}{3!}$

                         $$\begin{gathered} =\frac{{\mathrm{e}}^{{\displaystyle \frac{-4}{3}}} {\left({\displaystyle \frac{4}{3}}\right)}^3}{3!} \\ =\frac{{\mathrm{e}}^{{\displaystyle \frac{-4}{3}}}\times 64}{6\times 27} \\ =\frac{32{\mathrm{e}}^{{\displaystyle \frac{-4}{3}}}}{81} \end{gathered}$$