Answer
$Area\left( S \right) = 21\sqrt 2 \pi $

Work Step by Step
Since the surface of the part of the cone ${x^2} + {y^2} = {z^2}$ is located between the planes $z=2$ and $z=5$, we have the domain ${\cal D}$ given by
${\cal D} = \left\{ {\left( {x,y} \right):4 \le {x^2} + {y^2} \le 25} \right\}$
We can parametrize the surface by $G\left( {x,y} \right) = \left( {x,y, \pm \sqrt {{x^2} + {y^2}} } \right)$. Since $z$ is positive, we choose the parametrization $G\left( {x,y} \right) = \left( {x,y,\sqrt {{x^2} + {y^2}} } \right)$.
${{\bf{T}}_x} = \dfrac{{\partial G}}{{\partial x}} = \left( {1,0,\dfrac{x}{{\sqrt {{x^2} + {y^2}} }}} \right)$
${{\bf{T}}_y} = \dfrac{{\partial G}}{{\partial y}} = \left( {0,1,\dfrac{y}{{\sqrt {{x^2} + {y^2}} }}} \right)$
${\bf{N}}\left( {x,y} \right) = {{\bf{T}}_x} \times {{\bf{T}}_y} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&0&{\dfrac{x}{{\sqrt {{x^2} + {y^2}} }}}\\
0&1&{\dfrac{y}{{\sqrt {{x^2} + {y^2}} }}}
\end{array}} \right|$
${\bf{N}}\left( {x,y} \right) = - \dfrac{x}{{\sqrt {{x^2} + {y^2}} }}{\bf{i}} - \dfrac{y}{{\sqrt {{x^2} + {y^2}} }}{\bf{j}} + {\bf{k}}$
$||{\bf{N}}\left( {x,y} \right)|| = \sqrt {{{\left( { - \dfrac{x}{{\sqrt {{x^2} + {y^2}} }}} \right)}^2} + {{\left( { - \dfrac{y}{{\sqrt {{x^2} + {y^2}} }}} \right)}^2} + 1} $
$||{\bf{N}}\left( {x,y} \right)|| = \sqrt {\dfrac{{2{x^2} + 2{y^2}}}{{{x^2} + {y^2}}}} = \sqrt 2 $
By Theorem 1, the surface area of $S$ is
$Area\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} ||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$
$Area\left( S \right) = \sqrt 2 \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}x{\rm{d}}y$
Notice that $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}x{\rm{d}}y$ is the area of the annular region ${\cal D} = \left\{ {\left( {x,y} \right):4 \le {x^2} + {y^2} \le 25} \right\}$. So, we have
$Area\left( {\cal D} \right) = \pi \cdot{5^2} - \pi \cdot{2^2} = 21\pi $
Thus, the surface of the part of the cone is
$Area\left( S \right) = 21\sqrt 2 \pi $
