A pipe closed at end vibrating in the fourth overtone is in unison...

For a closed pipe, the frequency of the $p$-th overtone is given by $f_c = \dfrac{(2p+1)v}{4l_c}$.

For the fourth overtone ($p=4$), the frequency is $f_c = \dfrac{9v}{4l_c}$.

For an open pipe, the frequency of the $p$-th overtone is given by $f_o = \dfrac{(p+1)v}{2l_o}$.

For the fourth overtone ($p=4$), the frequency is $f_o = \dfrac{5v}{2l_o}$.

Since the two pipes are in unison, their frequencies are equal:

$\dfrac{9v}{4l_c} = \dfrac{5v}{2l_o}$

$\dfrac{l_c}{l_o} = \dfrac{9 \times 2}{4 \times 5} = \dfrac{18}{20} = \dfrac{9}{10}$

The ratio $l_c : l_o$ is $9 : 10$.

Answer: $9 : 10$