The required minimum is $5\sqrt{2}/6$: the circle of radius $5\sqrt{2}/6$, centered at $(5/6, 5/6)$ passes through exactly three lattice points, namely, $(0, 0)$, $(1, 2)$ and $(2, 1)$.

Next, consider a circle $\omega$ of radius at most $5\sqrt{2}/6$ passing through exactly three lattice points. We may and will assume that one of these lattice points is at the origin; let $(m, n)$ and $(p, q)$ be the other two.
Since the diameter of $\omega$ does not exceed $5\sqrt{2}/3$, the integers $m^2 + n^2$, $p^2 + q^2$ and $(m-p)^2 + (n-q)^2$ are all at most $5$.

If $m = 0$, then $p \neq 0$ by non-collinearity, and $n \neq 0$ must be even and $q = n/2$, for otherwise $(p, n-q)$ would be a fourth lattice point on $\omega$. Hence $n = \pm 2$ and $q = \pm 1$ (corresponding signs), and $|p|$ is either $1$ or $2$. The case $|p| = 1$ is ruled out by the fact that $(-p, q)$ would be a fourth lattice point on $\omega$, and the case $|p| = 2$ is ruled out by the fact that the radius of $\omega$ would be $5/4 > 5\sqrt{2}/6$.
Consequently, $m \neq 0$; similarly, $n \neq 0$, $p \neq 0$, $q \neq 0$, $m \neq p$, and $n \neq q$.
It then follows that $(m, n)$ and $(p, q)$ are consecutive vertices of the hexagon with vertices at $(2, 1)$, $(1, 2)$, $(-1, 1)$, $(-2, -1)$, $(-1, -2)$, $(1, -1)$ or of its reflection in one of the coordinate axes. The radius of $\omega$ is $5\sqrt{2}/6$ whatsoever the case.