Step 1: Recall the skew-symmetric shape.
A $3\times3$ skew-symmetric matrix has zeros on the diagonal, and each below-diagonal entry is the negative of the one above it. So only three entries $a_{12}, a_{13}, a_{23}$ are free.
Step 2: List the available values.
We must use the six numbers $a, -a, b, -b, c, -c$. The diagonal is forced to be $0$, and the lower triangle is forced once the upper triangle is chosen.
Step 3: Choose the first upper entry.
For $a_{12}$ we may pick any of the $6$ non-zero values. That uses one value; its negative automatically goes to $a_{21}$.
Step 4: Choose the second upper entry.
For $a_{13}$, the value just used and its negative are now taken, leaving $4$ choices.
Step 5: Choose the third upper entry.
For $a_{23}$, two more values are now used up, leaving $2$ choices.
Step 6: Multiply the choices.
Total $= 6 \times 4 \times 2 = 48$, which is option (D).
\[ \boxed{48} \]