Area of sector = $\dfrac{\theta}{360} \times \pi r^2 = 54\pi$, so $\dfrac{\theta}{360} \times (36)^2 = 54$, giving $\theta = 15°$.
Length of arc = $\dfrac{15}{360} \times 2\pi \times 36 = 3\pi$ cm.
Wait — re-checking: $\dfrac{\theta}{360} \times \pi \times 1296 = 54\pi \Rightarrow \theta = \dfrac{54 \times 360}{1296} = 15°$.
Arc length $= \dfrac{15}{360} \times 2\pi \times 36 = 3\pi$ cm.
Hmm, none match directly. Using area = $\dfrac{1}{2} \times l \times r$: $54\pi = \dfrac{1}{2} \times l \times 36 \Rightarrow l = 3\pi$ cm.
The correct answer is D: $3\pi$ cm — but since the closest listed option is (D) 3, the answer is D.
Use the relation: Area of sector = ½ × arc length × radius, i.e., $54\pi = \frac{1}{2} \times l \times 36$, giving $l = 3\pi$ cm. The options appear to omit $\pi$; option D ($3\pi$ cm) is correct. Alternatively, find $\theta = 15°$ from the area formula, then apply arc length $= \frac{\theta}{360} \times 2\pi r$. Both methods give $3\pi$ cm, corresponding to option D.