Answer
$F=-W k$
Work Step by Step
Divergence Theorem: $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV $
$div F=\dfrac{\partial p}{\partial x}+\dfrac{\partial q}{\partial y}+\dfrac{\partial r}{\partial z}$
As per the given problem, we have $\iint_S fc \cdot n dS=\iiint_Ediv (fc) dV$
$\implies \iint_S (fc) \cdot n dS=\iiint_E f( \nabla \cdot c) +(\nabla f) \cdot (c) dV$
$\implies \iint_S fc \cdot n dS=\iiint_E f(0) +(\nabla f) \cdot (c) dV$
This yields: $\iint_S fn \cdot c dS=\iiint_E (\nabla f) \cdot c dV$
and $\iint_S (f \cdot n) dS=\iiint_E (\nabla f) dV$
Also, $\nabla P=\dfrac{\partial (p)}{\partial x}i+\dfrac{\partial (p)}{\partial y}j+\dfrac{\partial (p)}{\partial z}k=\rho g k$
Since, $F=-\iiint_E (\nabla P) dV=-(\rho g k) \iiint_E dV$
Here, $\rho$ shows the density of the liquid and the integral $\iiint_E dV$ shows the volume of the solid.
But the weight of the liquid is: $W=\rho g v$
Thus, we get $F=-W k$
