= Solution
\b[The cobalt-60 mass is approximately $5.73\times10^{-8}\ \mathrm{kg}$.] The <nuclear decay constant> follows from the <half-life>, using a $365.25$-day year:
$$
\lambda=\frac{\ln2}{5.27(365.25)(86400)}
=4.16784\times10^{-9}\ \mathrm{s^{-1}}.
$$
Since <radioactive activity> is $A=\lambda N$, the parent population at certification is
$$
N=\frac{2.4\times10^9}{4.16784\times10^{-9}}
=5.75838\times10^{17}.
$$
Convert the supplied <atomic mass> into <molar mass>, $M=0.05993382\ \mathrm{kg\,mol^{-1}}$, and use the <Avogadro constant>:
$$
\boxed{m=\frac{NM}{N_A}
=\frac{(5.75838\times10^{17})(0.05993382)}
{6.02214076\times10^{23}}
=5.73088\times10^{-8}\ \mathrm{kg}.}
$$
This is about $57.3\ \mathrm{\mu g}$ of the specified <isotope>. It is not the total <mass> of the pellets or capsule, which can also contain other material.
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