If in a regular polygon the number of diagonals is 54, then the number of sides of this polygon is:

For a regular or even an irregular polygon, the total number of diagonals depends only on the total number of sides it possesses. The well-known formula giving this relation is stated first:

$$\text{Number of diagonals}= \frac{n(n-3)}{2},$$

where $$n$$ represents the number of sides of the polygon. This formula comes from the reasoning that each vertex can be joined to $$n-3$$ non-adjacent vertices to create diagonals, giving $$n(n-3)$$ such joins, and every diagonal is counted twice (once from each end), so we divide by $$2$$.

According to the question, the polygon has exactly $$54$$ diagonals. So we equate the formula to $$54$$:

$$\frac{n(n-3)}{2}=54.$$

To clear the fraction, we multiply both sides by $$2$$:

$$n(n-3)=108.$$

Next, we expand the left-hand side:

$$n^2-3n = 108.$$

Bringing all terms to one side gives a quadratic equation in standard form:

$$n^2-3n-108 = 0.$$

We now solve this quadratic. Using the quadratic formula $$n = \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$ for an equation $$ax^{2}+bx+c=0,$$ we have here $$a=1,\; b=-3,\; c=-108.$$ Substituting these values yields

$$n = \frac{-(-3) \pm \sqrt{(-3)^{2}-4(1)(-108)}}{2(1)}$$ $$= \frac{3 \pm \sqrt{9+432}}{2}$$ $$= \frac{3 \pm \sqrt{441}}{2}.$$

Since $$\sqrt{441}=21,$$ we arrive at two possible numerical results:

$$n = \frac{3 + 21}{2}= \frac{24}{2}=12,$$ $$n = \frac{3 - 21}{2}= \frac{-18}{2}=-9.$$

The second value $$-9$$ is not physically meaningful for the number of sides, because a polygon must have a positive whole number of sides. Therefore only

$$n = 12$$

is acceptable.

Hence, the correct answer is Option A.