For a certain organ pipe, the first three resonance frequencies are...

It is given,

${f}_{5} = 405Hz$.

As the first three resonance frequencies are in the ratio of $1: 3: 5$, therefore it must be a closed organ pipe.

The frequency formula in a closed organ pipe is given by $f = \dfrac{nv}{4l}$. So,

$\dfrac{5v}{4I} = 405Hz$

$⇒ I = \dfrac{405 \times 4}{5 \times 324} = 1m$