The square of the radius of the circle passing through the point $\left(6,-2\right)$ and  the equaitons of normals to the circle are $x+y=6$ and $x+2y=4$ is 

The given normals are $x+y=6$ and $x+2y=4$

The point of intersection of two normals is center of the cirlce 

Solve the equations  $x+y=6$ and $x+2y=4$

Subtract the first equation from the second equation 

$\begin{array}{rcl}x+2y-x-y& =& 4-6\\ y& =& -2\end{array}$

hence, $x=8$

Hence, the center of the circle i $\left(8,-2\right)$

Since the circle is passing throug the point $\left(6,-2\right)$, the radius of the circle is distance between the center  $\left(8,-2\right)$ and a point $\left(6,-2\right)$

Hence,  Radius of the circle is $\sqrt{\left|8-6\right|}=\sqrt{2}$

Therefore, square of the radius of the circle is $\boxed{2}$ units