Let the number of quarters in the first, second, third, and fourth piles be $ a $, $ b $, $ c $, and $ d $, respectively. We have the equations \begin{align*} \tag{1} a+b+c+d&=20\\ \tag{2} a&=b-3\\ \tag{3} b&=c+2\\ \tag{4} d&=2b \end{align*} We want to find the value of $ d $. We will express each of $ a $, $ b $, and $ c $ in terms of $ d $ and then substitute these equations into Equation (1) to find the value of $ d $. From Equation (4), we have $ b=d/2 $. From Equation (3), we have $ c=b-2 $. Since $ b=d/2 $, we can rewrite Equation (3) as $ c=d/2-2 $. We can substitute $ b=d/2 $ into Equation (2) to get $ a=d/2-3 $. Substituting $ b=d/2 $, $ c=d/2-2 $, and $ a=d/2-3 $ into Equation (1) to eliminate $ a $, $ b $, and $ c $, we get $ (d/2-3)+d/2+(d/2-2)+d=20 $, so $ d=10 $. Thus, there are $ \boxed{10} $ quarters in the fourth pile.