= Solution
Let $X$ be the original input, $U$ the output of the first channel, and $Y$ the final output. The <independent> erasure flag is observable from $Y$. The <mutual information> chain rule, conditioning on that flag, gives
$$
I(X;Y)=(1-\epsilon)I(X;U)+\epsilon\cdot0.
$$
The same uniform input maximizes $I(X;U)$ as in the <binary symmetric channel>. Thus
$$
\boxed{C=(1-\epsilon)(1-h(p))}.
$$
This scaling works because the erasure event is <independent> of both the input and the first channel noise.
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