Given that, one of the diameter of the circle ${\left(x-\sqrt{2}\right)}^2+{\left(y-3\sqrt{2}\right)}^2=6$ is a chord of another circle ${\left(x-2\sqrt{2}\right)}^2+{\left(y-2\sqrt{2}\right)}^2=r^2$.

Let, $C$ and $C_1$ be the centres of the circles ${\left(x-\sqrt{2}\right)}^2+{\left(y-3\sqrt{2}\right)}^2=6$ and ${\left(x-2\sqrt{2}\right)}^2+{\left(y-2\sqrt{2}\right)}^2=r^2$ respectively.

So, $C\left(\sqrt{2}, 3\sqrt{2}\right) \& C_1\left(2\sqrt{2}, 2\sqrt{2}\right)$

The position of the two circles as shown in figure.

Now, $C{C}_1=\sqrt{{\left(2\sqrt{2}-\sqrt{2}\right)}^2+ {\left(2\sqrt{2}-3\sqrt{2}\right)}^2}=\sqrt{2+2}=2$

It is clear from the above diagram, $∆AC{C}_1$ is a right angled triangle.$\Rightarrow r^2={\left(AC\right)}^2+{\left(C{C}_1\right)}^2={\left(\sqrt{6}\right)}^2+2^2$ $\Rightarrow r^2=10$