We call the $n \times n \times n$ cube the "large cube", and we call the $1 \times 1 \times 1$ cubes "unit cubes". The unit cubes that have exactly 0 gold faces are those unit cubes that are on the "inside" of the large cube.
In other words, these are the unit cubes none of whose faces form a part of any of the faces of the large cube.
These unit cubes form a cube that is $(n-2) \times(n-2) \times(n-2)$.
To see why this is true, imagine placing the original painted large cube on a table.
Each unit cube with at least one face that forms part of one of the outer faces (or outer layers) has paint on at least one face.
First, we remove the top and bottom layers of unit cubes. This creates a rectangular prism that is $n-2$ cubes high and still has a base that is $n \times n$.
Next, we can remove the left, right, front, and back faces.
This leaves a cube that is $(n-2) \times(n-2) \times(n-2)$.
Therefore, $(n-2)^{3}$ unit cubes have 0 gold faces.
The unit cubes that have exactly 1 gold face are those unit cubes that are on the outer faces of the large cube but do not touch the edges of the large cube.
Consider each of the six $n \times n$ faces of the large cube. Each is made up of $n^{2}$ unit cubes.
The unit cubes that have 1 gold face are those with at least one face that forms part of a face of the large cube, but do not share any edges with the edges of the large cube. Using a similar argument to above, we can see that these unit cubes form a $(n-2) \times(n-2)$ square.
There are thus $(n-2)^{2}$ cubes on each of the 6 faces that have 1 painted face, and so $6(n-2)^{2}$ cubes with 1 painted face.
We calculate the values of $(n-2)^{3}$ and $6(n-2)^{2}$ for each of the possible choices for $n$:
$$\begin{array}{c|c|c|c} Choice & n & (n-2)^{3} & 6(n-2)^{2} \\ \hline (A) & 7 & 125 & 150 \\ (B) & 8 & 216 & 216 \\ (C) & 9 & 343 & 294 \\ (D) & 10 & 512 & 384 \\ (E) & 4 & 8 & 24 \end{array}$$
From this information, the smallest possible value of $n$ when $(n-2)^{3}$ is larger than $6(n-2)^{2}$ must be $n=9$.
To see this in another way, we can ask the question "When is $(n-2)^{3}$ greater than $6(n-2)^{2}$?". Note that $(n-2)^{3}=(n-2) \times(n-2)^{2}$ and $6(n-2)^{2}=6 \times(n-2)^{2}$, and so $(n-2)^{3}$ is greater than $6(n-2)^{2}$ when $(n-2)$ is greater than 6, which is when $n$ is greater than 8.
The smallest positive integer value of $n$ for which this is true is $n=9$.