The electric flux ($$\Phi$$) through a closed surface, according to Gauss's law, is proportional to the enclosed charge $$Q$$, and is given by $$\Phi = \frac{Q}{\epsilon_0}$$, where $$\epsilon_0$$ is the vacuum permittivity. This is regardless of the shape of the surface, as long as it is closed and it encloses the charge fully.

In this case, both cubes $$C_1$$ and $$C_2$$ are closed surfaces, and they are concentric, meaning they share the same center. Each cube encloses a different charge, $$2Q$$ for $$C_1$$ and $$3Q$$ + $$2Q$$ = $$5Q$$ for $$C_2$$. According to Gauss's law, the electric flux through each cube is directly proportional to the enclosed charge.

For $$C_1$$, enclosing charge $$2Q$$: $$\Phi_{C_1} = \frac{2Q}{\epsilon_0}$$

For $$C_2$$, enclosing charge $$5Q$$: $$\Phi_{C_2} = \frac{5Q}{\epsilon_0}$$

Now, we want the ratio of electric flux through $$C_1$$ to that through $$C_2$$:

$$\frac{\Phi_{C_1}}{\Phi_{C_2}} = \frac{\frac{2Q}{\epsilon_0}}{\frac{5Q}{\epsilon_0}} = \frac{2Q}{5Q} = \frac{2}{5}$$

Therefore, the ratio of electric flux passing through $$C_1$$ to that passing through $$C_2$$ is $$\frac{2}{5}$$, which corresponds to option C.