Answer
$\iint_S f \cdot n dS=\iiint_E (\nabla f) dV$
Work Step by Step
Divergence Theorem: $\iiint_Ediv \overrightarrow{F}dV=\iint_S \overrightarrow{F}\cdot d\overrightarrow{S} $
where, $div F=\dfrac{\partial P}{\partial x}+\dfrac{\partial Q}{\partial y}+\dfrac{\partial R}{\partial z}$
$\iint_S fc \cdot n dS=\iiint_Ediv (fc) dV$
This implies that $\iint_S fc \cdot n dS=\iiint_E f( \nabla \cdot c) +(\nabla f) \cdot c dV$
and $\implies \iint_S fc \cdot n dS=\iiint_E f(0) +(\nabla f) \cdot c dV$
$\implies \iint_S fn \cdot c dS=\iiint_E (\nabla f) \cdot c dV$
and $\iint_S f \cdot n dS=\iiint_E (\nabla f) dV$
Hence the result has been verified.
