Step 1: Understand the flux calculation.
The flux \( \Phi \) through a surface is given by the surface integral of the dot product of the vector field and the normal vector to the surface:
\[
\Phi = \int_S \mathbf{F} \cdot \hat{n} \, dA
\]
where \( \mathbf{F} \) is the vector field, \( \hat{n} \) is the unit normal vector to the surface, and \( dA \) is the differential area element.
Step 2: Set up the integral.
The surface ABCD is a square in the \( xy \)-plane (since the problem does not give specific geometry, we'll assume it's in the \( xy \)-plane for simplicity). The normal vector \( \hat{n} \) is in the \( z \)-direction, and hence:
\[
\hat{n} = \hat{z}
\]
Therefore, the flux simplifies to:
\[
\Phi = \int_S (4yz) \, dA
\]
Since the surface is on the \( xy \)-plane, \( z = 0 \) on the surface. Thus, the flux becomes:
\[
\Phi = \int_S (4y \cdot 0) \, dA = 0
\]
Step 3: Conclusion.
The flux through the surface is zero, so the correct answer is 0.00.