Take the afternoon to begin at 12:00 noon. Then 1:20 pm is

$t=1\text{ hr }20\text{ min}=\tfrac43\text{ hours}, \qquad \lambda t = 5\cdot\tfrac43=\tfrac{20}{3}.$

Let $T$ be the waiting time to the first arrival; for a Poisson process $T\sim\operatorname{Exp}(\lambda)$ , and the fundamental link is

$\mathbb{P}(T>t)=\mathbb{P}\bigl(N(t)=0\bigr)=e^{-\lambda t}=e^{-20/3}.$

That quantity is option (B), and the numerical agreement of $\lambda t$ with the printed exponent confirms the intended time window and rate.

Note — the question and its options do not match. The event "the first message arrives by 1:20 pm" is $\{T\le\tfrac43\}$ , whose probability is the complement

$\mathbb{P}\bigl(T\le\tfrac43\bigr)=1-e^{-20/3}\approx0.99873 ,$

which is not among the choices. Option (B) is $\mathbb{P}(T>\tfrac43)\approx0.00127$ , i.e. the probability that the first message has not arrived by 1:20 pm. Option (A) is $\mathbb{P}\bigl(N(\tfrac43)=1\bigr)$ and option (D) is $\mathbb{P}\bigl(N(\tfrac13)=1\bigr)$ . Of the four, only (B) corresponds to a natural quantity at the stated rate and horizon, and it is taken here as the intended key — the stem should have read "arrives after 1:20 pm".