= Solution
For equal allocation, let $n$ be the required <sample size> per arm, $p_C=0.0045$, $p_T=0.003$ and $\Delta=0.0015$. In the standard <sample size for comparing two proportions> calculation, a <two-sided test> at 5% with 80% <statistical power> requires
$$
\Delta\sqrt n\simeq z_{0.975}\sqrt{2\bar p(1-\bar p)}+z_{0.8}\sqrt{p_C(1-p_C)+p_T(1-p_T)},\qquad \bar p=0.00375.
$$
Using $z_{0.975}=1.95996$ and $z_{0.8}=0.84162$ gives $n\simeq26063.64$, which is rounded upward for actual recruitment. Thus
$$
\boxed{N_{\rm total}\simeq52128,\quad\text{about }52.1\text{ thousand participants, or about }53\text{ thousand when planning in whole thousands}.}
$$
Since the risks are small, dropping the $(1-p)$ corrections and using $1.96+0.84$ gives $N_{\rm total}\simeq2(2.8)^2(0.0075)/(0.0015)^2=52266.7$, conventionally about 52.3 thousand. These are slightly different approximations to the same planning requirement. Add an allowance if outcome ascertainment is incomplete or the allocation is clustered. A prespecified <one-sided test> would need a different, smaller <sample size>; the calculation here uses the conventional two-sided interpretation consistently with part (b).
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