Bulk modulus of water is $2.3\times {10}^9\mathrm{N}/{\mathrm{m}}^2$. Taking average density of water $\mathrm{\rho }={10}^3\mathrm{kg}/{\mathrm{m}}^3$ , find increase in density at a depth of 1km in kg /m³ . Take $\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2$.

Pressure increases with depth of a liquid At a depth ‘h’ below the water surface increase in pressure is given by $\mathrm{\Delta p}=\mathrm{\rho gh}$
We get  $\mathrm{\Delta \rho }=\mathrm{\rho }\frac{\mathrm{\Delta \rho }}{\mathrm{B}}$
$\mathrm{\Delta \rho }=\frac{\mathrm{\rho }\left(\mathrm{\rho gh}\right)}{\mathrm{B}}=\frac{{\mathrm{\rho }}^2\mathrm{gh}}{\mathrm{B}}$
Substituting the values we have, 
$\mathrm{\Delta \rho }=\frac{{\left({10}^3\right)}^2\left(10\right)\left({10}^3\right)}{2.3\times {10}^9}=4.35\mathrm{kg}/{\mathrm{m}}^3$