We apply the linear momentum balance equation page 206 of Munson 5th edition
$\begin{align}\frac{\partial }{\partial t}\int_{CV} \mathbf{V} \rho d V + \int_S \mathbf{V} \rho \mathbf{V} \cdot \mathbf{n} d...$
where $\sum \mathbf{F}$ is the net force on the contents of the control volume. We choose as our control volume a cube such that the stream enters orthogonally through one side of the cube and the opposite side of the cube is at the wall. This is a steady flow, so the first term in the equation above is zero. We assume that gravity can be neglected, so the net force on the control volume is due only to the wall. Also, because the stream enters orthogonally through one side of the cube, $\mathbf{V} \cdot \mathbf{n} = -V$ for the part of the surface integral involving the entering stream -- the minus sign arises because the stream's velocity is antiparallel to the outward pointing normal vector.

We consider only the x-component of the equation above. The velocity of the stream entering has only an x-component, and the streams leaving have no x-component. So, the x-component of the equation becomes

$\setcounter{equation}{1}\begin{align}\int_S V \rho \mathbf{V} \cdot \mathbf{n} dA = -\int_S \mathbf{V} \rho V dA = F_{wall}\e...$
where $\mathbf{V}$ here denotes the velocity of the stream as it enters the control volume. We assume that $\rho$ and $\mathbf{V}$ are constants. So, the equation becomes
$\setcounter{equation}{2}\begin{align}-V^2 \rho A = F_{wall}\end{align}$
The minus sign indicates that the force is in the minus x-direction, which it should be. Assuming that it is the magnitude of the force that is sought, then answer (A) is correct.

Solution to 2001 Problem 57