The correct answer is $\frac{4}{3}$.
The momentum correction factor is a dimensionless quantity that is used to account for the fact that the velocity is not uniform across the cross-sectional area of a pipe. The momentum correction factor is defined as follows:
$$f = \frac{2}{3} \left( 1 + \frac{2}{3} \frac{A_1}{A} \right)$$
where $A_1$ is the area of the half of the cross-sectional area where the velocity is zero and $A$ is the total cross-sectional area of the pipe.
In this case, the velocity is zero over half of the cross-sectional area and is uniform over the remaining half. Therefore, $A_1 = \frac{1}{2} A$ and $A = 2 A_1$. Substituting these values into the equation for the momentum correction factor, we get:
$$f = \frac{2}{3} \left( 1 + \frac{2}{3} \frac{A_1}{A} \right) = \frac{2}{3} \left( 1 + \frac{2}{3} \frac{1}{2} \right) = \frac{4}{3}$$
Therefore, the momentum correction factor is $\frac{4}{3}$.
The other options are incorrect because they do not account for the fact that the velocity is not uniform across the cross-sectional area of the pipe.