Answer
The volume decreases by $6.71\times 10^{-6}~m^3$
Work Step by Step
We can find the the change in volume:
$\frac{\Delta V}{V_0} = -\frac{1}{B}~\Delta P$
$\Delta V = -\frac{1}{B}~\Delta P~V_0$
$\Delta V = -\left(\frac{1}{60\times 10^9~Pa}\right)~(1.75\times 10^6~Pa)(0.230~m^3)$
$\Delta V = -6.71\times 10^{-6}~m^3$
The volume decreases by $6.71\times 10^{-6}~m^3$.
