Let $ a $ be the number of coins Amy has, $ b $ the number Ben has, $ c $ the number Carl has, and $ d $ the number Debbie has. We can use the information in the problem to create the following system of linear equations: \begin{align*} 3a &= b \\ 3b &= c \\ \frac{2}{3}c &= d \\ a \cdot b \cdot c \cdot d &= 162 \end{align*} We know that $ b = 3a $. Also, $ c = 3b $, so $ c $ must equal $ 9a $. Then, $ d = \frac{2}{3}c $, so $ d = 6a $. Substituting these quantities into our product gives $ a \cdot 3a \cdot 9a \cdot 6a = 162, $ which simplifies to $ 162a^4 = 162, $ or $ a^4 = 1, $ so $ a = 1 $. Given this value, we can find that $ b = 3 $, $ c = 9 $, and $ d = 6 $. Thus, the four children have $ 1 + 3 + 9 + 6 = \boxed{19} $ coins all together.