Solution:

To calculate the flux \(\Phi_E\) of the electric field \(\vec{E}\) through the square, we use the formula:

\[
\Phi_E = \vec{E} \cdot \vec{A}
\]

Here:
- \(\vec{E} = 2 \times 10^3 \, \text{N/C}\) (along the positive x-axis),
- \(\vec{A}\) is the area vector, perpendicular to the plane of the square.

Since the square lies in the yz-plane, its area vector points along the x-axis (same direction as \(\vec{E}\)), and its magnitude is the area of the square:

\[
\text{Side of square} = 10 \, \text{cm} = 0.1 \, \text{m}, \quad \text{Area} = (0.1)^2 = 0.01 \, \text{m}^2
\]

The flux is:

\[
\Phi_E = |\vec{E}| \cdot |\vec{A}| \cdot \cos\theta
\]

Here, \(\theta = 0^\circ\) (since \(\vec{E}\) is parallel to \(\vec{A}\)):

\[
\Phi_E = (2 \times 10^3) \cdot 0.01 \cdot \cos 0^\circ = 20 \, \text{N·m}^2/\text{C}
\]

Thus, the flux is:

\[
{20 \, \text{N·m}^2/\text{C}}
\]