Answer
$R_o = 7.53 \times 10^{16} \mathrm{s}^{-1}$
Work Step by Step
$$
\lambda=(\ln 2) / T_{1 / 2}=(\ln 2) /(78 \mathrm{h})=8.89 \times 10^{-3} \mathrm{h}^{-1}
$$
The mass $m$ of a single atom of gallium is
$$
m=(67 \mathrm{u})\left(1.661 \times 10^{-24} \mathrm{g} / \mathrm{u}\right)=1.113 \times 10^{-22} \mathrm{g}
$$
If $M$ is the mass of the sample then $N_{o}=$ $M / m .$
$$
N_{o}=(3.4 \mathrm{g}) /\left(1.113 \times 10^{-22} \mathrm{g}\right)=3.05 \times 10^{22}
$$
Thus,
$$
R_{o}= \lambda N_o = \left(8.89 \times 10^{-3} \mathrm{h}^{-1}\right)\left(3.05 \times 10^{22}\right)=2.71 \times 10^{20} \mathrm{h}^{-1}=\boxed{7.53 \times 10^{16} \mathrm{s}^{-1}}
$$
