Answer
(b)
Work Step by Step
We have $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {{\bf{F}}\cdot{\bf{n}}} \right){\rm{d}}S$.
If ${\bf{F}}\left( P \right) = {\bf{n}}\left( P \right)$ at each point on $S$, then
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {{\bf{F}}\cdot{\bf{n}}} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {{\bf{n}}\cdot{\bf{n}}} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} 1{\rm{d}}S$
Since $Area\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\rm{d}}S$, so the answer is (b).
