Let $A$ be the event that the item is sent to A, $B$ be the event that the product is sent to B, and $D$ be the event that the product is defective. We are given $$\begin{align} \Pr(A)&=.3\\\ \Pr(B)&=.7\\\ \Pr(D|A)&=.05\\\ \Pr(D|B)&=.04 \end{align}$$
Now, $$\Pr(A\cap D)=\Pr(D|A)\Pr(A)=.05\cdot.3=.015$$
You have computed $\Pr(D)\Pr(A)$ which assumes that $A$ and $D$ are independent. This cannot be the case; A gets a higher percentage of defective items than B does.