Given the use of $\beta$, I assume the parameterization of the exponential distribution in terms of mean.
As you should know, the MLE is the parameter that maximizes $P(\text{obs} | \text{parameter})$.
As the observations are i.i.d, we can simply multiply the probabilities of the individual observations together to get the overall likelihood function. As these are binary variables, this is actually a thinly disguised binomial distribution. The overall likelihood is then just $P \propto p^k (1-p)^{n-k}$. The CDF of the exponential distribution gives $p = P(X_i \gt 5 | \beta) = \exp(-5/\beta)$.
Choosing the $\beta$ that maximizes $P$ can be done by choosing the $p$ that does.
For $k$ successes (i.e. observations > 5) in $n$ trials, the MLE for a binomial is a standard exercise, with the result that $p = k/n$.
Letting $p = \exp(-5/\beta) = k/n$ gives $\beta = -5/log(k/n)$.
If all observations are $\gt 5$ (i.e. $k=n$), this has no maximum over the real numbers; in a loose sense, $\beta=\infty$ would maximizes this. A more sensible answer would require going beyond the MLE framework with regularization or a Bayesian approach.
