Density of water, $\rho  = {10^3}\,kg\,{m^{ - 3}}$

Compressibility of water, $K = 45.4 \times {10^{ - 11}}P{a^{ - 1}}$

$\frac{{\Delta V}}{V} = ?$

Excess pressure at the bottom, $\Delta P = \rho gd$

$ = {10^3} \times 10 \times 2700 = 27 \times {10^6}Pa$

$We\,know,\,B = \frac{{\Delta P}}{{\left( {\Delta V/V} \right)}}$

$\left( {\frac{{\Delta V}}{V}} \right) = \frac{{\Delta P}}{B} = K.\Delta P$              $\left( {K = \frac{1}{B}} \right)$

$ = 45.4 \times {10^{ - 11}} \times 27 \times {10^6} = 1.2 \times {10^{ - 2}}$