Solution:

To find the charge enclosed, we use Gauss's law:

\[
Q = \oint \vec{E} \cdot d\vec{A} = \epsilon_0 \frac{dV}{dx} \cdot A
\]

Given \( V(x) = 3 - 2x^3 \), we compute the electric field:

\[
E_x = -\frac{dV}{dx} = 6x^2
\]

Now, for a cube with side 1m and one vertex at the origin, the flux through the cube's face on the x-axis is:

\[
\text{Flux} = E_x \cdot A = 6x^2 \cdot 1 = 6
\]

Since only the face along the x-axis contributes to the flux, the total charge enclosed is:

\[
Q = \epsilon_0 \cdot 6
\]

Thus, \( n = 6 \).