There are two approaches, through the Poisson and through the associated exponential.
The number of events in two days is a Poisson random variable $Y$ with parameter $2\lambda$. The probability that $Y=0$ is $e^{-2\lambda}$.
Or else the waiting time $W$ is an exponentially distributed random variable with parameter $2$. The probability that $W\gt 2$ is $$\int_2^\infty \lambda e^{-\lambda t}\,dt.$$ Integrate (or remember the cdf of the exponential). We get $e^{-2\lambda}$.