The answer is $5$ for all $n \ge 1000$. If $n \ge 200$ we have $n^2+n+200 < n^2+2n+1 = (n+1)^2$, so $n < \sqrt{n^2+n+200} < n+1$ and $\lfloor \sqrt{n^2+n+200} \rfloor = n$. It also follows that $\sqrt{n^2+n+200}$ is not an integer, moreover $\sqrt{n^2+n+200}$ is irrational. Let $k$ be the first digit of $\sqrt{n^2+n+200}$ after the decimal point, $0 \le k \le 9$. Then $n + \frac{k}{10} < \sqrt{n^2+n+200} < n + \frac{k+1}{10}$, or

$$10n + k < 10\sqrt{n^2 + n + 200} < 10n + (k + 1).$$

The inequalities are strict as $\sqrt{n^2+n+200}$ is irrational. Squaring and simplification gives
$$20nk + k^2 < 100n + 20000 < 20n(k + 1) + (k + 1)^2.$$

The left inequality implies $20n(k-5) < 20000$, $n(k-5) < 1000$. Given $n \ge 1000$, we see that $k \le 5$. Otherwise $k-5 \ge 1$ and $n(k-5) \ge n \ge 1000$. The right inequality can be rewritten as $20000 < 20n(k-4) + (k+1)^2$. Hence $20000 < 20n(k-4) + 10^2$ because $k \le 9$; thus $1000 < n(k-4) + 5$. So $n(k-4) > 0$ which implies $k \ge 5$. Now $k \le 5$ and $k \ge 5$ lead to $k=5$.