Answer is (B)
$20$ request in $1$ hour. So we can expect $15$ request in $45$ minutes...
So, $\lambda = 15$ (expected value)
Poisson distribution formula$: f(x, \lambda) = p(X = x) = \dfrac{e^{-\lambda}*\lambda^x}{x!}$
$\text{Prob (1 request)} + \text{Prob (3 requests)} + \text{Prob (5 requests)}$
$\quad= p(1; 15) + p(3; 15) + p(5; 15)$
$\quad= {6.9} \times 10^{3} \times e ^ {-15}$
$\quad = {6.9}\times 10^{3}\times e^{5}\times e^{-20} $
$\quad= {1.02}\times {10^6}\times e^{-20}.$