Answer
$$16 \ \text { fissions / day. }
$$
Work Step by Step
Using Eq. $42-20$ and adapting Eq.$42-21$ to this sample, the number of fission events per second is
$$
\begin{aligned} R_{\text {fision }} &=\frac{N \ln 2}{{T_{1 / 2}}_{ fision }}=\frac{M_{\text {sam }} N_{A} \ln 2}{M_{\mathrm{U}} {T_{1 / 2}}_{ fision }} \\ &=\frac{(1.0 \mathrm{g})\left(6.02 \times 10^{23} / \mathrm{mol}\right) \ln 2}{(235 \mathrm{g} / \mathrm{mol})\left(3.0 \times 10^{17} \mathrm{y}\right)(365 \mathrm{d} / \mathrm{y})}=16 \text { fissions / day. } \end{aligned}
$$
