Given a Poisson process with an average rate of $\lambda = 6.0$ defaults per year, the time until the next default follows an exponential distribution. The probability that the next default occurs within time $t$ (in years) is given by the cumulative distribution function of the exponential distribution: $P(T \le t) = 1 - e^{-\lambda t}$.

For one month, $t = 1/12$ years. Thus, the probability is $1 - e^{-6.0 \times (1/12)} = 1 - e^{-0.5} \approx 1 - 0.6065 = 0.3935$, or `$39.35`%$.