In Poisson distribution :
Mean = Variance as n is large and p is small
And we know:
$Variance=E(X^2)−[E(X)]^2 $
$⇒E(X^2)=[E(X)]^2+Variance = λ^2 + λ$
Similarly, $ E(Y^2)=[E(Y)]^2+Variance = λ^2 + λ$
$ E[2XY] = 2 * E[X] * E[Y] $
$ E(X−Y)^2 = E[X^2+2XY+Y^2] = E[X^2] + E[2XY] + E[Y^2]$
$= λ^2 + λ - 2.λ.λ+ λ^2 + λ) $
$ = 2λ$
So B is correct.
Ref: https://gateoverflow.in/118513/gate2017-2-48