We evaluate this in cases:
Case 1
$x0$. When $x>0$ we are going to have $|x|>0\implies x>0$ and when $-x>0$ we are going to have $|x|>0\implies -x>0$. Therefore we have $x=|2x-(60-2x)|$.
$x=|2x-60+2x|\implies x=|4x-60|$
Subcase 1 $30>x>15$
When $30>x>15$ we are going to have $4x-60>0$. When this happens, we can express $|4x-60|$ as $4x-60$.
Therefore we get $x=4x-60\implies -3x=-60\implies x=20$. We check if $x=20$ is in the domain of the numbers that we put into this subcase, and it is, since $30>20>15$. Therefore $20$ is one possible solution.
Subcase 2 $x30$
When $x>30$, $60-2x<0$. When $x<0$ we can express this in the form $-x$. Therefore we have $-(60-2x)=2x-60$. This makes sure that this is positive, since we just took the negative of a negative to get a positive. Therefore we have
$x=|2x-(2x-60)|$
$x=|2x-2x+60|$
$x=|60|$
$x=60$
We have now evaluated all the cases, and found the solution to be $\{60,12,20\}$ which have a sum of $\boxed{\textbf{(C)}\ 92}$