Answer
$$
\lambda=7.46 \times 10^{-5} \mathrm{s}^{-1}
$$
$$N=1.19 \times 10^{15}$$
Work Step by Step
We use $N=R / \lambda .$ If $T_{1 / 2}$ this is half-life, so disintegration constant will be:
$$
\lambda=(\ln 2) / T_{12}=(\ln 2) /(2.58 \mathrm{h})=0.269 \mathrm{h}^{-1}=7.46 \times 10^{-5} \mathrm{s}^{-1}
$$
so $$N=\left(8.88 \times 10^{10} \mathrm{s}^{-1}\right) /\left(7.46 \times 10^{-5} \mathrm{s}^{-1}\right)=1.19 \times 10^{15}$$
