Answer
$f ∘ G =\frac{1}{|x-2|}$
Therefore the domain is
$x\ne2$
Work Step by Step
$f(x) = |x|$
$G(x) = \frac{1}{(x-2)}$
$f ∘ G = f(G(x)) = f( \frac{1}{(x-2)}) = | \frac{1}{(x-2)}|=\frac{1}{|x-2|}$
Therefore the domain is
$|x-2|\ne0$
$x-2\ne 0$
$x\ne2$