Answer
$-1.4\times 10^{-6}m^3$
Work Step by Step
We can find the required volume decrease as follows:
$P-P_{\circ}=\rho gh$
We plug in the known values to obtain:
$P-P_{\circ}=(1\times 10^3Kg/m^3)(9.8m/s^2)(10.9\times 10^3m)$
$P-P_{\circ}=1.068\times 10^8Pa$
Now $\Delta V=\frac{-V_{\circ}\Delta P}{B}$
We plug in the known values to obtain:
$\Delta V=\frac{-(17.66\times 10^{-4}m^3)(1.068\times 10^8Pa)}{14\times 10^{10}N/m^2}$
$\Delta V=-1.4\times 10^{-6}m^3$
