The eccentricity of the hyperbola which passes through the points $(3,0)$ and $(3 \sqrt{2}, 2)$ is

Given that the hyperbola passes through the points $(3, 0)$ and $(3\sqrt{2}, 2)$, we take the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Substituting point $(3, 0)$ yields $\frac{9}{a^2} = 1$, thus $a^2 = 9$.

With $a^2$ known, inserting point $(3\sqrt{2}, 2)$ gives $\frac{18}{9} - \frac{4}{b^2} = 1$, which simplifies to $2 - \frac{4}{b^2} = 1$ and leads to $b^2 = 4$.

The eccentricity follows from $b^2 = a^2(e^2 - 1)$. Substituting known values, $4 = 9(e^2 - 1)$, so $e^2 = \frac{13}{9}$ and $e = \frac{\sqrt{13}}{3}$.

The correct choice is C.