Answer
$2.65\times10^{-4}m^{3}$
Work Step by Step
When the system is in equilibrium.
Buoyant force + paperweight in water = actual weight of the paper
$F_{B}+4.3\space N=6.9\space N=>F_{B}=2.6\space N$
According to the principle of Archimedes', we can write
$F_{B}=V_{w}\rho_{w}g=>V_{w}=\frac{F_{B}}{\rho_{w}g}$
Let's plug known values into this equation.
$V_{w}=\frac{2.6\space N}{1000\space kg/m^{3}\times9.8\space m/s^{2}}=2.65\times10^{-4}m^{3}$ (volume of water)
