The unit cubes with $3$ yellow faces are at the corners of the cube. There are therefore $8$ of them. The unit cubes with $2$ yellow faces are located along the edges of the cube. There are $12$ edges and each has $n – 2$ unit cubes with $2$ yellow faces for a total of $12(n – 2)$ unit cubes with $2$ yellow faces. The unit cubes with only one yellow face are in the interior of the $6$ faces of the cube. There are $(n – 2)^2$ such cubes in each face for a total of

$6(n – 2)^2$ unit cubes with one yellow face.

Finally, the unit cubes with no yellow faces form a smaller cube inside the big cube. There are $(n – 2)^3$ of these.

The problem gives us two equations:

(# unit cubes with 1 yellow face) = $2\cdot$(# unit cubes with 2 yellow faces)

in other words $6(n – 2)^2 = 2\cdot 12(n – 2)$.

And

(# unit cubes with no yellow faces) = $8\cdot$ (# unit cubes with 3 yellow faces)

or $(n – 2)^3 = 8\cdot8$.

Now the first equation is a quadratic with the two roots: $2$ and $6$. Only the root $6$ satisfies the second equation, so $n = 6$ is the answer.