Consider the length of the tube be $L$

The fundamental frequency for the open organ pipe is,

$f=\frac{V}{2 \mathrm{L}}$

The figure below represents the $60 \%$ of tube is dipped
vertically in water.

The length of the closed organ pipe is,

$l^{\text{'}}=\frac{40}{100} L$

$l^{\text{'}}=\frac{2 \mathrm{L}}{5}$

The fundamental frequency of air column is,

$f^{\text{'}}=\frac{V}{4{\mathrm{l}}^{\text{'}}}$

Substitute the values in the above equation.

$f^{\text{'}}=\frac{V}{4 \left(\frac{2 \mathrm{L}}{5}\right)}$

$f^{\text{'}}=\frac{5}{4}\left(\frac{V}{2 \mathrm{L}}\right)$

$f^{\text{'}}=\frac{5}{4}f$