= Solution
For equal independent samples, take $p_B=0.0002$, $p_C=0.0006$ and $\bar p=0.0004$ in the two-proportion formula, with two-sided $\alpha=0.05$ and power $1-\beta=0.80$. It gives
$$
n\simeq\frac{\left[1.95996\sqrt{2(0.0004)(0.9996)}
+0.84162\sqrt{0.0002(0.9998)+0.0006(0.9994)}\right]^2}
{(0.0004)^2}
=39227.52.
$$
Thus the usual normal planning estimate is
$$
\boxed{\text{about }3.92\text{ blocks of 10,000 per supplier,
or }4\text{ whole blocks each}.}
$$
The combined approximate design therefore checks about eight blocks, with about eight positive batches from B and twenty-four from C at the rounded design.
Those expected counts are low enough for test discreteness to matter. If a genuinely exact two-sided <Fisher exact test> is specified, the null allocation of the positive batches between two equal-size suppliers is <hypergeometric>, conditional on their total. Under the proposed alternative, its power is obtained by summing the independent binomial <probabilities> over the exact rejection region:
$$
\operatorname{Power}(n)=\sum_{b,c:\,p_F(b,c)\leq0.05}
\Pr\{\operatorname{Bin}(n,0.0002)=b\}
\Pr\{\operatorname{Bin}(n,0.0006)=c\}.
$$
An independent calculation of the <exact power of Fisher's exact test> gives about $0.7750$ at $n=40000$ and $0.8728$ at $n=50000$. Consequently \b[five whole 10,000-batch blocks per supplier suffice for at least 80% power with this conservative exact test]. Four blocks are the intended normal-approximation answer, not an exact 80% guarantee irrespective of the chosen test. The distinction is a property of rare-event discreteness, not a change in the proposed effect size.
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