Step 1: Understanding the Concept
For a closed pipe the fundamental frequency is \(n=\dfrac{v}{4(L+e)}\), where \(e\approx0.3d\) is the end correction and \(d\) the diameter.
Step 2: Apply the Condition
Both pipes have the same length \(L\), but A is wider so \(e_{A}>e_{B}\).
Step 3: Compare
Effective length of A is larger, so \(n_{A}\lt n_{B}\).
Final Answer:
The wider pipe A has the lower frequency, so nA is less than nB.
\[ \boxed{\text{(C) }n_{A}\lt n_{B}} \]