First overtone of open pipe $$ = {{{v_2}} \over {{L_2}}}$$

First overtone of closed pipe at one end $$ = {{3v} \over {4L}}$$

As per question,

$${{3V} \over {4L}} = {{{V_2}} \over L}$$

$$ \Rightarrow \sqrt {{B \over {{\rho _1}}}} \,.\,{3 \over {4L}} = \sqrt {{B \over {{\rho _2}}}} \,.\,{1 \over {{L_2}}}$$ ($$\because$$ $$V = \sqrt {{B \over \rho }} $$)

$$ \Rightarrow {L_2} = {{4L} \over 3}\sqrt {{{{\rho _1}} \over {{\rho _2}}}} $$ ..... (i)

According to question, the length of the open pipe is

$${x \over 3}L\sqrt {{{{\rho _1}} \over {{\rho _2}}}} $$ ..... (ii)

Comparing Eqs. (i) and (ii), we get

$$x = 4$$