Solution:

The electric flux is given by:

\[
\Phi = \vec{E} \cdot \vec{A} = E A \cos\theta,
\]

where:
- \(E = 2 \times 10^3 \, \text{V/m}\),
- \(A = \text{area of the coil} = 10 \, \text{cm} \times 20 \, \text{cm} = 0.1 \, \text{m} \times 0.2 \, \text{m} = 0.02 \, \text{m}^2\),
- \(\theta = 0^\circ\) (field is perpendicular to the coil, as the coil is in the \(xy\)-plane and the field is along \(\hat{k}\)).

Substitute values:
\[
\Phi = (2 \times 10^3) \cdot (0.02) \cdot \cos(0^\circ),
\]
\[
\Phi = 40 \, \text{V·m}.
\]

Thus, the electric flux is:

\[
40 \, \text{V·m}
\]