The 4^th overtone of a closed organ pipe is same as that of 3^rd overtone of an open pipe. The ratio of the length of the closed pipe to the length of the open pipe is :

n^th overtone of a closed pipe is $(2n+1)\frac{V}{4\ell }$

n^th overtone of an open pipe is $(n+1)\frac{V}{2\ell }$
4^th overtone of a closed pipe is $\frac{9\mathrm{V}}{4{\ell }_{\mathrm{c}}}$

3^rd overtone of an open pipe is $\frac{4\mathrm{V}}{2{\ell }_0}$
According to the given condition,

$\frac{9\mathrm{V}}{4{\ell }_{\mathrm{c}}}=\frac{4\mathrm{V}}{2{\ell }_0}$

$\frac{{\ell }_c}{{\ell }_0}=\frac{9}{8}$