Step 1: Poisson Probability Formula
\[
P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}
\]
Given \( \lambda = 5 \), we find:
\[
P(X3) = P(0) + P(1) + P(2)
\]
\[
P(0) = \frac{e^{-5} 5^0}{0!} = e^{-5}
\]
\[
P(1) = \frac{e^{-5} 5^1}{1!} = 5 e^{-5}
\]
\[
P(2) = \frac{e^{-5} 5^2}{2!} = \frac{25}{2} e^{-5}
\]
Step 2: Summation
\[
P(X3) = e^{-5} + 5 e^{-5} + \frac{25}{2} e^{-5}
\]
\[
= \left(1 + 5 + \frac{25}{2} \right) e^{-5}
\]
\[
= \frac{37}{2} e^{-5}
\]
Thus, the correct answer is \( \frac{37}{2 e^{-5}} \).