The second overtone of an open organ pipe has the same frequency as the first overtone of a closed pipe which is $L$ meter long. The length of the open pipe will be

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

Frequency of second overtone of open organ pipe is given as, $f_1=\frac{3v}{2L\text{'}}$, where $v$ is speed of sound in air.

Frequency of first overtone of closed pipe is given as, $f_2=\frac{3v}{4L}$.

According to question, $f_1=f_2$

$\frac{3v}{2L\text{'}}=\frac{3v}{4L}\Rightarrow L\text{'}=2L$.

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