If cube X has an edge of length e and cube Y has an edge of length e +

These kinds of questions (Variables in the Answer Choices - VIACs) can be answered algebraically or using the INPUT-OUTPUT approach.
The posters above have solved the question algebraically, so let's use the INPUT-OUTPUT approach.

Let's say e = 3

Cube X
Each edge has length 3
Since each face of the cube is a SQUARE, the area of ONE face = 3² = 9
So, the area of all SIX faces = (6)(9) = 54

Cube Y
Each edge has length 3 + 1. So, each edge has length 4
Since each face of the cube is a SQUARE, the area of ONE face = 4² = 16
So, the area of all SIX faces = (6)(16) = 96

How much greater is the total surface area of cube Y than that of cube X?
Difference = 96 - 54 = 42

So, when we INPUT e = 3, the answer to the question (i.e., the OUTPUT) is 42

Now we'll examine each answer choice, to see which one yields and output of 42 when we replace e with 3

We get:
A. 12(3) + 6 = 42 Great! Keep.
B. 6(3 + 1)² = 24. We want an output of 42. ELIMINATE B
C. 3 + 1 = 4. We want an output of 42. ELIMINATE C
D. 3 = 3. We want an output of 42. ELIMINATE D
E. 1 = 1. We want an output of 42. ELIMINATE E

Answer: A

Cheers,
Brent