= Solution
The <central limit theorem> gives $\sqrt n(\widehat\theta_n^{\mathrm{MC}}-\theta)\Rightarrow N(0,1)$, so, with $\Phi$ the <cumulative distribution function> of the <standard normal distribution>,
$$
\mathbb P\!\left(\left|\widehat\theta_n^{\mathrm{MC}}-\theta\right|<0.01\right)
\approx 2\Phi(0.01\sqrt n)-1.
$$
The error has two tails. Using the supplied $2.58$ <quantile>, whose two-sided coverage is approximately $0.9902$, gives
$$
0.01\sqrt n\geq2.58,
\qquad
\boxed{N:=n=\left\lceil\frac{2.58^2}{0.01^2}\right\rceil=66{,}564.}
$$
The supplied $2.33$ <quantile> would be appropriate for a one-sided $99\%$ calculation; it would give only approximately $98.02\%$ two-sided coverage. Using the exact $0.995$ <quantile> instead of $2.58$ would give $66{,}349$ draws; we consistently use the paper's rounded value $N=66{,}564$ below.
For routine <Monte Carlo integration>, the <central limit theorem> calculation is preferable as a practical budget: it uses about one fifteenth as many draws as the <Chebyshev's inequality> bound. However, its coverage is approximate. Finite <variance> alone supplies no universal rate of convergence to the <normal distribution>; an integrand with large <skewness> or a <heavy-tailed distribution> can make the approximation poor even at a large sample count. If an unconditional finite-sample $99\%$ guarantee is required under only the stated assumptions, use the <Chebyshev's inequality> budget.
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