The second overtone of an open organ pipe has the same frequency as...

Let the length of open organ pipe be $L'$.

Frequency of second overtone of open organ pipe is given as, ${f}_{1} = \dfrac{3v}{2L '}$, where $v$ is speed of sound in air.

Frequency of first overtone of closed pipe is given as, ${f}_{2} = \dfrac{3v}{4L}$.

According to question, ${f}_{1} = {f}_{2}$

$\dfrac{3v}{2L '} = \dfrac{3v}{4L} ⇒ L ' = 2L$.

Hence, length of open organ pipe must be twice the length of closed organ pipe.