Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = 0$
Work Step by Step
Since ${\bf{F}}$ is a unit vector field so, ${\bf{F}}\cdot{\bf{n}} = \cos \theta $.
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{S_1}}^{} \left( {{\bf{F}}\cdot{\bf{n}}} \right){\rm{d}}S + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{S_2}}^{} \left( {{\bf{F}}\cdot{\bf{n}}} \right){\rm{d}}S + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{S_3}}^{} \left( {{\bf{F}}\cdot{\bf{n}}} \right){\rm{d}}S$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{S_1}}^{} \cos 85^\circ {\rm{d}}S + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{S_2}}^{} \cos 90^\circ {\rm{d}}S + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{S_3}}^{} \cos 95^\circ {\rm{d}}S$
Since $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{S_1}}^{} {\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{S_2}}^{} {\rm{d}}S + \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{{S_3}}^{} {\rm{d}}S = 0.2$, so
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = 0.2\left( {\cos 85^\circ + \cos 90^\circ + \cos 95^\circ } \right) = 0$
