The closed and open organ pipe have same length and when they are vibrating simultaneously in first overtone produce $3$ beats. The length of open pipe is made $\left(\dfrac{1}{3}\right)^{rd}$ and closed pipe is made $3$ times the original, the number of beats produced will be {neglect end correction}

Let the original length of both pipes be $L$ and the speed of sound be $v$.

The frequency of the first overtone of the open pipe is $f_{o1} = \dfrac{2v}{2L} = \dfrac{v}{L}$.

The frequency of the first overtone of the closed pipe is $f_{c1} = \dfrac{3v}{4L}$.

Given that they produce $3$ beats per second:

$f_{o1} - f_{c1} = 3$

$\dfrac{v}{L} - \dfrac{3v}{4L} = 3$

$\dfrac{v}{4L} = 3 \Rightarrow \dfrac{v}{L} = 12$

When the length of the open pipe is made $\dfrac{L}{3}$, its new fundamental frequency is:

$f_{o}' = \dfrac{v}{2 \left(\dfrac{L}{3}\right)} = \dfrac{3v}{2L} = \dfrac{3}{2} \times 12 = 18$

When the length of the closed pipe is made $3L$, its new fundamental frequency is:

$f_{c}' = \dfrac{v}{4(3L)} = \dfrac{v}{12L} = \dfrac{1}{12} \times 12 = 1$

The new number of beats produced is:

$f_{o}' - f_{c}' = 18 - 1 = 17$

Answer: $17$