Answer
$18.37\%$
Work Step by Step
Please see the attached image first.
The weight of the displaced Mercury is $W_{m}=m_{m}g=\rho_{m}V_{sub}g$
By Archimedes' principle, $W_{m}$ is equal in magnitude to the buoyancy force, which balances gravity when the slab is in equilibrium. So,
$m_{m}g=\rho_{water}V_{sub}g$
$1700\space kg=V_{m}\times13.69\times10^{3}kg/m^{3}$
$0.124\space m^{3}=V_{m}$
We know that the,
$V_{slab}=0.3\space m\times1.5\space m\times1.5\space m=0.675\space m^{3}$
percentage of the slab’s volume was below $=\frac{V_{m}}{V_{slab}}\times100\%$
the surface of the mercury$(V\%)$
$V\%=\frac{0.124m^{3}}{0.675m^{3}}\times100\%=18.37\%$
