Answer
0.2 m
Work Step by Step
Please see the attached image first.
From figure (1), by considering the equilibrium of the system we can write
$F_{B1}=mg$
According to the Principle of Archimedes', we can write,
$V_{w}\rho_{w}g=mg$
$\frac{1}{3}H^{3}\rho_{w}=m-(1)$
From figure (2), by considering the equilibrium of the system we can write
$F_{B2}=mg+m_{w}g$
According to the Principle of Archimedes', we can write,
$V_{b}\rho_{w}g=mg+V_{w}\rho_{w}g$
(1)=>
$\rho_{w}H^{3}=\frac{1}{3}H^{3}\rho_{w}+\rho_{w}H^{2}h=>h=\frac{2}{3}H=\frac{2}{3}(0.3\space m)=0.2\space m$
Depth of the water in the box = 0.2 m
