Consider the equation of the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
It is given that the centre of the hyperbola is $\left(0,0\right)$ and its transverse axis is along $x$-axis and its length is $12$.
$2 a=12$
$a=6$

Substitute this value in the equation of a hyperbola.
$\frac{x^2}{6^2}-\frac{y^2}{b^2}=1$
The above equation passes through the point $\left( 8,2\right)$
$\frac{8^2}{6^2}-\frac{2^2}{b^2}=1$
$\frac{64}{36}-\frac{4}{b^2}=1$
After simplification,
$b=\frac{6}{\sqrt{7}}$
The equation of hyperbola becomes,

$\frac{x^2}{6^2}-\frac{y^2}{{\left(\frac{6}{\sqrt{7}}\right)}^2}=1$
$\frac{x^2}{36}-\frac{7y^2}{36}=1$
The eccentricity of a hyperbola is,

$e=\sqrt{1+\frac{a^2}{b^2}}$

$e=\sqrt{1+\frac{6^2}{{\left({\displaystyle \frac{6}{\sqrt{7}}}\right)}^2}}$

$e=\sqrt{\frac{8}{7}}$

$e=\frac{2\sqrt{2}}{\sqrt{7}}$