The correct answer is $\boxed{\frac{9}{4}}$.
The correction factor for kinetic energy is given by:
$$K = \frac{1}{2} \int_0^1 v^2(x) \, dx$$
where $v(x)$ is the velocity as a function of position $x$. In this case, $v(x) = 0$ for $x < \frac{1}{3}$ and $v(x) = v_0$ for $x \ge \frac{1}{3}$. Therefore, the integral becomes:
$$K = \frac{1}{2} \int_0^{\frac{1}{3}} 0 \, dx + \frac{1}{2} \int_{\frac{1}{3}}^1 v_0^2 \, dx = \frac{v_0^2}{12}$$
The correction factor is then given by:
$$\frac{K}{K_0} = \frac{\frac{v_0^2}{12}}{\frac{1}{2} v_0^2} = \frac{9}{4}$$
where $K_0$ is the kinetic energy if the velocity were uniform over the entire cross-section.
Option A is incorrect because it is the correction factor for a case where the velocity is zero over $\frac{2}{3}$ of the cross-section. Option B is incorrect because it is the correction factor for a case where the velocity is uniform over the entire cross-section. Option C is incorrect because it is the correction factor for a case where the velocity is zero over $\frac{1}{6}$ of the cross-section.