A Poisson variate x is such that  $P\left(x=2\right)=9P\left(x=4\right)+90.P\left(x=6\right)$ then mean  and standard deviation are

$P\left(X=2\right)=9P\left(X=4\right)+90P\left(X=6\right)$

$e^{-\lambda }\frac{{\lambda }^2}{2!}=9.e^{-\lambda }\frac{{\lambda }^4}{4!}+90e^{-\lambda }\frac{{\lambda }^6}{6!}$

$1=9.\frac{{\lambda }^2}{12}+90\frac{{\lambda }^4}{360}$

$1=\frac{3{\lambda }^2}{4}+\frac{{\lambda }^4}{4}$

$\Rightarrow {\lambda }^4+3{\lambda }^2-4=0$

$\Rightarrow {\lambda }^2=1\Rightarrow \lambda =1$

Mean=1 & S.D=1