= Solution
Use equal allocation and a conventional two-sided 5% comparison of independent proportions. The assumed risks are $p_C=0.005$ and $p_I=0.005(1-0.30)=0.0035$, giving target difference $\delta=0.0015$. Put $\overline p=(p_C+p_I)/2=0.00425$ and let $m$ be the number per arm.
The null critical difference is approximately $z_{0.975}\sqrt{2\overline p(1-\overline p)/m}$. Under the alternative, the difference has standard deviation $\sqrt{[p_C(1-p_C)+p_I(1-p_I)]/m}$. For 80% <statistical power>, the target difference must exceed the critical difference by $z_{0.8}$ alternative standard deviations. Solving gives the <sample size for comparing two proportions>
$$
m\simeq\frac{\left[z_{0.975}\sqrt{2\overline p(1-\overline p)}+z_{0.8}\sqrt{p_C(1-p_C)+p_I(1-p_I)}\right]^2}{\delta^2}.
$$
This is about $29{,}500$ per arm, or $59{,}000$ altogether. Rounding for planning:
$$
\boxed{\text{about 60 thousand participants in total, roughly 30 thousand per arm}.}
$$
The rare-event Poisson approximation gives a similar answer. This assumes the stated baseline risk applies, no endpoint loss, and no attenuation by <treatment contamination>. The two-sided convention is explicit: a one-sided 5% test would require fewer participants.
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