Frequency of $p$ th overtone of open organ pipe
$=(p+1) \frac{v}{2 L}$...(i)
where, $L=$ length of organ pipe and that of closed organ pipe is
$=(2 p+1) \frac{V}{4 L}$...(ii)
So, the ratio of $p$ th overtone of open to closed
$=\frac{(p+1) \frac{v}{2 L}}{(2 p+1) \frac{v}{4 L}}$ (using Eqs. (i) and (ii)
$=\frac{2(p+1)}{(2 p+1)}$