= Solution
Apply <Chebyshev's inequality> to the <Monte Carlo estimator>:
$$
\mathbb P\!\left(\left|\widehat\theta_n^{\mathrm{MC}}-\theta\right|\geq0.01\right)
\leq\frac{\operatorname{Var}(\widehat\theta_n^{\mathrm{MC}})}{0.01^2}
=\frac{10^4}{n}.
$$
Thus $10^4/n\leq0.01$ suffices, giving
$$
\boxed{n=1{,}000{,}000.}
$$
This is a finite-sample <probability> guarantee for every target satisfying the given <variance> assumption. It is a sufficient sample count supplied by the inequality, rather than a claim that every such target actually needs a million draws.
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