We know, for closed pipe,

$${f_n} = {{nv} \over {4l}},\,n = 1,3,5,7$$

for open pipe, $${f_n} = {{nv} \over {2L}},\,n = 1,2,3,4$$

So, 9$^{th}$ harmonic of closed pipe $$ = {{9{v_1}} \over {4{l_1}}}$$

4$^{th}$ harmonic of open pipe $$ = {{4{v_2}} \over {2{l_2}}} = {{2{v_2}} \over {{l_2}}}$$

$\therefore$ $${{9{v_1}} \over {4{l_1}}} = {{2{v_2}} \over {{l_2}}} \Rightarrow {{{l_2}} \over {{l_1}}} = {8 \over 9}{{{v_2}} \over {{v_1}}}$$

We know, $$v = \sqrt {{\beta \over \rho }} $$

So, $${{{v_2}} \over {{v_1}}} = \sqrt {{{{\rho _1}} \over {{\rho _2}}}} $$ (for same $\beta$)

hence, $${{{l_2}} \over {{l_1}}} = {8 \over 9}\sqrt {{{{\rho _1}} \over {{\rho _2}}}} = {8 \over 9}\sqrt {{1 \over {16}}} = {8 \over 9} \times {1 \over 4}$$

$$ \Rightarrow {l_2} = {2 \over 9} \times {l_1} \Rightarrow {l_2} = {2 \over 9} \times 10 = {{20} \over 9}cm$$