1. Given electric field

$\vec E=(2\hat i+4\hat j+6\hat k)\times 10^3\,\text{N/C}$

So the components are:

$E_x=2\times10^3,\quad E_y=4\times10^3,\quad E_z=6\times10^3\,\text{N/C}$

2. Identify the relevant component for flux

The surface is parallel to the $x$-$z$ plane.

Hence its area vector is perpendicular to the $x$-$z$ plane, i.e. along the $y$-axis. So only the $y$-component of electric field contributes to flux.

$E_\perp=E_y=4\times10^3\,\text{N/C}$

3. Use the formula for electric flux

$\Phi=\vec E\cdot \vec A = E_\perp A$

Given:

$\Phi=6.0\,\text{N m}^2\text{/C}$

Thus,

$A=\frac{\Phi}{E_\perp} = \frac{6.0}{4\times10^3}\,\text{m}^2$

$A=1.5\times10^{-3}\,\text{m}^2$

4. Convert into $\text{cm}^2$

Since

$1\,\text{m}^2=10^4\,\text{cm}^2$

therefore,

$A=1.5\times10^{-3}\times10^4\,\text{cm}^2$

$A=15\,\text{cm}^2$

5. Final answer

$\boxed{15}$

This matches the stored correct answer.