To calculate the area of the surface through which the electric flux is determined, we use the given electric field and the specified surface orientation. The electric field is expressed as $\overrightarrow{\mathrm{E}} = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times 10^3 \, \mathrm{N/C}$.

Since the surface is parallel to the $x-z$ plane, its area vector $\overrightarrow{\mathrm{A}}$ is directed along the $y$-axis, making it $\mathrm{~A} \hat{\mathrm{j}}$.

The electric flux $\phi$ through the surface is given by:

$ \phi = \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{~A}} = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times 10^3 \cdot \mathrm{~A} \hat{\mathrm{j}} $

Calculating the dot product, the only component contributing to the flux is the $y$-component:

$ \phi = (4 \times 10^3) \mathrm{~A} $

Given that the electric flux $\phi$ is $6.0 \, \mathrm{Nm}^2/\mathrm{C}$, we can equate and solve for $\mathrm{A}$:

$ 6 = 4 \times 10^3 \mathrm{~A} $

$ \mathrm{A} = \frac{6}{4 \times 10^3} = 1.5 \times 10^{-3} \, \mathrm{m}^2 $

Converting this area from square meters to square centimeters:

$ \mathrm{A} = 1.5 \times 10^{-3} \, \mathrm{m}^2 = 15 \, \mathrm{cm}^2 $