The electric flux through a surface is given by the formula:

$$\Phi = \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}} = |\overrightarrow{\mathrm{E}}||\overrightarrow{\mathrm{A}}|\cos\theta$$

where $$\overrightarrow{\mathrm{E}}$$ is the electric field, $$\overrightarrow{\mathrm{A}}$$ is the area vector (with magnitude equal to the area of the surface and direction perpendicular to the surface, defined by the unit vector $$\hat{n}$$), and $$\theta$$ is the angle between $$\overrightarrow{\mathrm{E}}$$ and $$\overrightarrow{\mathrm{A}}$$. However, when using unit vectors to describe the directions of $$\overrightarrow{\mathrm{E}}$$ and $$\hat{n}$$, the dot product can be used to simplify the calculation as follows:

$$\Phi = \overrightarrow{\mathrm{E}} \cdot \left(\overrightarrow{\mathrm{A}}\right) = (\overrightarrow{\mathrm{E}} \cdot \hat{n})A$$

Given that $$\overrightarrow{\mathrm{E}}=\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}$$ and $$\hat{n}=\left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)$$, and the area $$A = 4 \mathrm{m}^2$$, we can substitute them into our formula. Note that since $$\overrightarrow{\mathrm{A}} = A\hat{n}$$, the magnitude of the area vector is the area of the surface itself. First, let's find $$\overrightarrow{\mathrm{E}} \cdot \hat{n}$$:

$$\overrightarrow{\mathrm{E}} \cdot \hat{n} = \left(\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}\right) \cdot \left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)$$

To compute the dot product, we multiply corresponding components and then add them up:

$$\left(\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}\right) \cdot \left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right) = \frac{1}{6}(2\cdot2 + 6\cdot1 + 8\cdot1)$$

$$= \frac{1}{6}(4 + 6 + 8) = \frac{18}{6} = 3$$

Then, the electric flux through the surface is:

$$\Phi = (\overrightarrow{\mathrm{E}} \cdot \hat{n})A = 3 \times 4 \mathrm{m}^2$$

$$\Phi = 12 \mathrm{Vm}$$

So, the electric flux for that surface is $$12 \mathrm{Vm}$$.