 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 proportionsThis is about $29,500$ per arm, or $59,000$ altogether. Rounding for planning: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.