I did in the following way,

For any time span probability of there being x transmission attempts is

$$\frac{A^xe^{-A}}{x!}$$

Where A is the average no of transmission attempt in that time span.

In that question time duration is $2T$ . And we know that channel load G is the average transmission attempt in T time duration. So, in 2T time the average transmission attempts will be 2G.

SO, plugging A = 2G in the above formula,

$$\frac{(2G)^xe^{-2G}}{x!}$$

For the given condition , No transmission attempts is 2T => x = 0

Probability = $\frac{(2G)^0e^{-2G}}{0!}$ = 0.5

=> G = 0.3465

Now for poisson Mean of average is = Np

=> p = 0.003465 (very low)