Two closed organ pipes A and B have the same length. A is wider than B. They resonate in the fundamental mode at frequencies n_A and n_B respectively, then

In closed organ pipe, first resonance occurs at $\lambda /4$. So, in fundamental mode of vibration of organ pipe $\lambda /4=(l+0.3d)$ where 0.3 d is necessary end correction, then frequency of  vibration,  $n=\frac{v}{\lambda }=\frac{v}{4(l+0.3d)}$

As l is same, wide pipe A will resonate at a lower frequency.

So,  $n_A<n_B$.