Answer
$f ∘ g ∘ G =|(\frac{1}{(x-2)})^2 - 4|$
Therefore the domain is
$x\ne2$
Work Step by Step
$f(x) = |x|$
$G(x) = \frac{1}{(x-2)}$
$g(x) = x^2-4$
$f ∘ g ∘ G = f(g(G(x))) = f( g(\frac{1}{(x-2)})) = f((\frac{1}{(x-2)})^2 - 4)=|(\frac{1}{(x-2)})^2 - 4|$
Therefore the domain is
$x-2\ne0$
$x\ne2$