Historical data:
Am I right to think about this question from the perspective of Poisson Distribution? I think I have to find something like $P(X\le 3)$, but I don't know how to find $\lambda$, the mean number of occurrences in this case.
In this case, we need to make sure there are less than $3$ patients at a specific time $t$.
How do you find the number of dentists so less than a specific number of patients are waiting at specific time t?
First of all get an estimate of the inter-arrival rate from the dataset. Calculate interarrival times as $I_i=a_i-a_{i-1}$ where $a_i$'s are arrival times and $a_0=0$. Then $\hat\lambda=\frac{n}{\sum I_i}.$ Now let's say that only $m$ many patients are allowed to wait and you want $n$ many dentists. So, at a time $t$ you want inter-arrival less than or equal to $m+n$ between time $(t-10,t]$, otherwise, the clinic has to return some patients. So, you want $N_t-N_{t-10}\left(\stackrel{d}{=}N_{10}\right)\leq m+n.$ Now $N_t-N_{t-10}\sim \textbf{Poi}(10\hat\lambda)$ and as @TonyK mentioned in the comment that there is no way you can make sure that with probability one. However, you can do it for expectation or quantiles.
For an example, if you want on an average how many dentists you need then you answer should be $E(N_t-N_{t-10}\leq m+n)\Rightarrow 10\hat\lambda\leq m+n\Rightarrow n\geq 10\hat\lambda-m.$
Again if you want how many doctors you need such that with probability at most $p$ the clinic don't have to return a patient, then find the smallest non-negative integer $k_0$ such that $P(N_{10}>k_0)\leq p.$ Then the answer should be $n\geq k_0-m.$