The first overtone frequency of an open organ pipe is equal to the...

The fundamental frequency of the closed organ pipe is given by, ${f}_{0} = \dfrac{v}{4{l}_{c}}$.

The frequency of the first overtone of a open organ pipe is given by, ${f}_{1} = 2 \times \dfrac{v}{2{l}_{o}}$.

Since,

${f}_{0} = {f}_{1} ⇒ \dfrac{v}{4{l}_{c}} = \dfrac{v}{{l}_{o}} ⇒ {l}_{o} = 4 \times 20 = 80cm$.