Answer
Please see the figure attached.
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\rm{d}}S \approx 0.0424$

Work Step by Step
The small patch (green) is illustrated in the figure attached.
From Exercise 9, we get $G\left( {\theta ,\phi } \right) = \left( {\cos \theta \sin \phi ,\sin \theta \sin \phi ,\cos \phi } \right)$. So,
${{\bf{T}}_\theta } = \frac{{\partial G}}{{\partial \theta }} = \left( { - \sin \theta \sin \phi ,\cos \theta \sin \phi ,0} \right)$
${{\bf{T}}_\phi } = \frac{{\partial G}}{{\partial \phi }} = \left( {\cos \theta \cos \phi ,\sin \theta \cos \phi , - \sin \phi } \right)$
${\bf{N}}\left( {\frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{4}} \right) = {{\bf{T}}_\theta } \times {{\bf{T}}_\phi } = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{ - \frac{1}{2}\sqrt 2 }&0&0\\
0&{\frac{1}{2}\sqrt 2 }&{ - \frac{1}{2}\sqrt 2 }
\end{array}} \right|$
${\bf{N}}\left( {\frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{4}} \right) = - \frac{1}{2}{\bf{j}} - \frac{1}{2}{\bf{k}}$
$||{\bf{N}}\left( {\frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{4}} \right)|| = \sqrt {0 + {{\left( { - \frac{1}{2}} \right)}^2} + {{\left( { - \frac{1}{2}} \right)}^2}} = \frac{1}{2}\sqrt 2 $
We estimate the area of the small patch $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\rm{d}}S$ by
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\rm{d}}S \approx ||{\bf{N}}\left( {\frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{4}} \right)||\Delta \theta \Delta \phi $
Since $\frac{\pi }{2} - 0.15 \le \theta \le \frac{\pi }{2} + 0.15$ and $\frac{\pi }{4} - 0.1 \le \phi \le \frac{\pi }{4} + 0.1$, so $\Delta \theta = 0.3$ and $\Delta \phi = 0.2$.
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\rm{d}}S \approx \frac{1}{2}\sqrt 2 \cdot 0.3 \cdot 0.2 = 0.0424$