Solution verification: Calculating expected value of sum of three unusual dice
To elaborate JMoravitz's comment, let $X$, $Y$ and $Z$ denote the outcome of three dice respectively (renameing the sum as $W$). Then what you want is $E(X+Y+Z)$, which by linearity of the expectation: $$ E(X+Y+Z) = E(X)+E(Y)+E(Z)\;, $$ which can be found by $$ E(X) = \frac16(1+3+5+7+9+10) = \frac{35}{6}, $$ $$ E(Y) = \frac13\cdot 2+\frac12\cdot 3+\frac16\cdot 1=\frac{14}{6}, $$ and $$ E(Z) = \frac13\cdot 2+\frac23\cdot 4=\frac{10}{3}\;. $$
[Added:]
Remark. This is actually connected to your attempt as follows (without using linearity). $$ \begin{align} E(W) &= \sum_{w\in A}wP(X+Y+Z=w) \end{align} $$ But for each $w\in A$, the event $\{X+Y+Z=w\}$ can be written as a disjoint union as $$ \{X+Y+Z=w\} = \bigcup_{w_1+w_2+w_3=w\\w_k\in D_k, k=1,2,3}(X = w_1,Y=w_2,Z =w_3) $$ Thus, $$ \begin{align} E(W) &= \sum_{w_k\in D_k}(w_1+w_2+w_3)P(X = w_1)P(Y = w_2)P(Z=w_3)\\ &=\sum_{w_k\in D_k}w_1P(X = w_1)P(Y = w_2)P(Z=w_3)\\ &+\sum_{w_k\in D_k}w_2P(X = w_1)P(Y = w_2)P(Z=w_3)\\ &+\sum_{w_k\in D_k}w_3P(X = w_1)P(Y = w_2)P(Z=w_3)\\ &= \sum_{w_1\in D_1}w_1P(X=w_1) +\sum_{w_2\in D_2}w_2P(Y=w_2) +\sum_{w_3\in D_3}w_3P(Z=w_3) \\ &= E(X)+E(Y)+E(Z)\;. \end{align} $$
$$E[X]=\frac{1+3+5+7+9+10}{6}+\frac{1+2+2+3+3+3}{6}+\frac{2+2+4+4+4+4}{6}=\frac{69}{6}=11.5$$
Let me illustrate the suggestion made by JMoravitz in the comments. Define $X_1$, $X_2$, and $X_3$ as the random variables whose outcomes are the roll of the first, second, and third dice, respectively. Then $$X = X_1 + X_2 + X_3,$$ so $$E(X) = E(X_1 + X_2 + X_3) = E(X_1) + E(X_2) + E(X_3)$$ by linearity of expectation. We have: $$E(X_1) = \sum_x x P(X_1 = x) = \frac{1}{6} \left(1 + 3 + 5 + 7 + 9 + 10 \right) = \frac{35}{6}.$$ $$E(X_2) = \sum_x x P(X_2 = x) = 1 \cdot \frac{1}{6} + 2 \cdot \frac{2}{6} + 3 \cdot \frac{3}{6} = \frac{14}{6}.$$ $$E(X_3) = \sum_x x P(X_3 = x) = 2 \cdot \frac{2}{6} + 4 \cdot \frac{4}{6} = \frac{20}{6}.$$ Hence: $$E(X) = \frac{35 + 14 + 20}{6} = \frac{69}{6} = 11.5.$$