We define $f(x) = x^2 - 3x + 13$ and note that $f(x) = (x-3)x + 13 = 13 - x(3-x)$ from which we easily discover that $f(3-x) = f(x)$ for all $x$. Calculating $f(0) = 13$ and $f(1) = 11$, we see that the smallest prime $p$ that divides $f(n)$ for some $n$ is at most $11$. We may finish in two different ways.

Way 1. Because $f(n) \equiv f(m) \pmod{p}$ when $n \equiv m \pmod{p}$, it is sufficient to calculate $f(n) \pmod{p}$ for $p$ consecutive values of $n$. We only need to do this for $p \in \{2, 3, 5, 7\}$ in order to determine whether $11$ is the smallest prime we are looking for. Because $f(3-n) = f(n)$, we have $f(2) = f(1) = 11$, $f(3) = f(0) = 13$ and $f(4) = f(-1) = 17$. Finally, $f(-2) = 23$ and we see that $f(n) \ge 11$ is a prime for each of the seven consecutive values $n = -2, -1, 0, 1, 2, 3, 4$. Hence, the smallest prime number that divides $f(n)$ for at least one $n$ is $p = 11$.

Way 2. For each $p \in \{2, 3, 5, 7\}$ we discuss $f(n) \pmod{p}$ separately.
$$f(n) = n^2 - 3n + 13 \equiv n^2 - n + 1 \equiv 1 \pmod{2}$$
because $n^2 \equiv n \pmod{2}$ for all integers $n$ by Fermat's Little Theorem.
$$f(n) = n^2 - 3n + 13 \equiv n^2 - 2 \not\equiv 0 \pmod{3}$$
because squares of integers can only be congruent to $0$ or $1$ modulo $3$.
$$f(n) = n^2 - 3n + 13 \equiv n^2 + 2n + 3 \equiv (n+1)^2 + 2 \not\equiv 0 \pmod{5}$$
because squares of integers can only be congruent to $0$ or $\pm 1$ modulo $5$.
$$f(n) = n^2 - 3n + 13 \equiv n^2 + 4n + 6 \equiv (n+2)^2 + 2 \not\equiv 0 \pmod{7}$$
because squares of integers can only be congruent to $0$, $1$, $2$ or $4$ modulo $7$. Hence, for no integer $n$ is $f(n)$ divisible by $2$, $3$, $5$ or $7$. The smallest prime number that divides $f(n)$ for at least one $n$ therefore is $p = 11$.