Answer
$0.126\ m^{3}$
Work Step by Step
The density of the casting equals
$\rho_{c}= \displaystyle \frac{m_{casting}}{V_{casting}}= \displaystyle \frac{m_{iron}}{V_{iron}+V_{cavities}}$
so,
$ \displaystyle \frac{V_{iron}+V_{cavities.}}{m_{iron}}=\frac{1}{\rho_{c}} $
$V_{iron}+V_{cavities}=\displaystyle \frac{m_{iron}}{\rho_{c}} $
$V_{cavities}=\displaystyle \frac{m_{iron}}{\rho_{c}} -V_{iron}\qquad (*)$
The mass of the iron equals the mass of the (iron+cavities),
In air, $m_{iron}g=6000N \displaystyle \Rightarrow m_{iron}=\frac{6000N}{g}$,
The volume of the iron is $V=\displaystyle \frac{m_{iron}}{\rho_{iron}}=\frac{6000N}{(9.8m/s^{2})(7.87\times 10^{3}\frac{kg}{m^{3}})}$
In air, we have
$\rho_{c}\cdot V_{casting}\cdot g=6000N$
In water, the buoyant force equals the difference in weights (air/water)
$\rho_{w}\cdot V_{casting}\cdot
g=2000N$
Dividing the last two equations,
$\displaystyle \frac{\rho_{c}}{\rho_{w}}=3.00 \displaystyle \quad\Rightarrow\quad\rho_{c}=3.00\rho_{w}=3000\frac{kg}{m^{3}}$
Insert these values into $(*)$
$V_{cavities}=\displaystyle \frac{\frac{6000N}{(9.8m/s^{2})}}{3000\frac{kg}{m^{3}}} -\frac{6000N}{(9.8m/s^{2})(7.87\times 10^{3}\frac{kg}{m^{3}})}=0.126m^{3}$
