Answer
The density increases by a fraction of 0.005 which is a percentage of 0.50%
Work Step by Step
We can find the gauge pressure at a depth of 1.0 km:
$P_g = \rho~g~h$
$P_g = (1025~kg/m^3)(9.80~m/s^2)(1000~m)$
$P_g = 1.00\times 10^7~Pa$
We can find the fraction that the volume changes:
$\frac{\Delta V}{V_0} = -\frac{1}{B}~\Delta P$
$\frac{\Delta V}{V_0} = -\frac{1}{2.0\times 10^9~Pa}~(1.00\times 10^7~Pa)$
$\frac{\Delta V}{V_0} = -5.0\times 10^{-3}$
We can find the new volume $V'$ in terms of the original volume $V_0$:
$V' = (1-5.0\times 10^{-3})~V_0 = 0.995~V_0$
We can find the new density $\rho'$ in terms of the original density $\rho_0$:
$\rho' = \frac{Mass}{V'}$
$\rho' = \frac{Mass}{0.995~V_0}$
$\rho' = 1.005\times \frac{Mass}{V_0}$
$\rho' = 1.005~\rho_0$
The density increases by a fraction of 0.005 which is a percentage of 0.50%
