When open pipe is closed from one end then third overtone of closed pipe is higher in frequency by 150Hz than second overtone of open pipe. The fundamental frequency of open end pipe will be (Neglect end correction)

Let the fundamental frequency of the open pipe be $f_o = \dfrac{v}{2L}$.

The frequency of the second overtone of the open pipe corresponds to the third harmonic, so $f_{o2} = 3f_o$.

When the pipe is closed at one end, its fundamental frequency becomes $f_c = \dfrac{v}{4L} = \dfrac{f_o}{2}$.

The frequency of the third overtone of the closed pipe corresponds to the seventh harmonic, so $f_{c3} = 7f_c = 7\left(\dfrac{f_o}{2}\right) = 3.5f_o$.

Given that the third overtone of the closed pipe is higher in frequency by $150\text{ Hz}$ than the second overtone of the open pipe:

$f_{c3} - f_{o2} = 150$

$3.5f_o - 3f_o = 150$

$0.5f_o = 150$

$f_o = 300\text{ Hz}$

Answer: $300\ \text{Hz}$