Answer
$$ \approx 5.04 \times 10^{18} \text { nuclei}
$$
Work Step by Step
We adapt Eq. $42-21:$
$$
N_{\mathrm{Pu}}=\left(\frac{0.002 \mathrm{g}}{239 \mathrm{g} / \mathrm{mol}}\right)\left(6.02 \times 10^{23} \text { nuclei/mol }\right) \approx 5.04 \times 10^{18} \text { nuclei. }
$$
