It is given that,

$2P(X=1)=5P(X=5)+2P(X=3)$,

The formula for Poisson $P\left(X=r\right)=\frac{e^{-\lambda }{\lambda }^r}{r!}$.

$\Rightarrow \frac{2e^{-\lambda }\lambda }{1!}=\frac{5e^{-\lambda }{\lambda }^5}{5!}+\frac{2e^{-\lambda }{\lambda }^3}{3!}$

$\Rightarrow 2=\frac{{\lambda }^4}{24}+\frac{{\lambda }^2}{3}$

$\Rightarrow 2=\frac{3{\lambda }^4+24{\lambda }^2}{72}$

$\Rightarrow 144=3{\lambda }^4+24{\lambda }^2$

$\Rightarrow {\lambda }^4+8{\lambda }^2-48=0$

Consider ${\lambda }^{\mathit{2}}\mathit{=}x$,

$\Rightarrow x^2+8x-48=0$

$\Rightarrow (x+12)(x-4)=0$

$x=4$ and $x\ne -12$

The standard deviation is,

$x={\lambda }^2=4$ and $\sqrt{\lambda }=\sqrt{2}$