1. Decay scheme from ${}_{92}^{238}U \to {}_{82}^{206}Pb$

To go from mass number $238$ to $206$, the decrease is $238-206=32$ Each $\alpha$-particle emission decreases mass number by $4$, so number of $\alpha$ particles emitted per uranium nucleus is $\frac{32}{4}=8$

Check atomic number:

• After $8\alpha$ emissions, atomic number decreases by $16$: $92-16=76$
• Final atomic number is $82$, so it must increase by $6$. Each $\beta^-$ emission increases atomic number by $1$, so there are $6\beta^-$ emissions.

Thus, each ${}_{92}^{238}U$ nucleus ultimately emits $8\alpha$ particles.

2. Initial number of ${}_{92}^{238}U$ atoms

Given mass of uranium: $68\times 10^{-6}\text{ g}$

Molar mass of ${}_{92}^{238}U \approx 238\text{ g mol}^{-1}$

So initial moles are $n=\frac{68\times 10^{-6}}{238}$

Initial number of atoms: $N_0=\frac{68\times 10^{-6}}{238}N_A$

Using $N_A=6.022\times 10^{23}$, $N_0=\frac{68\times 10^{-6}}{238}\times 6.022\times 10^{23}$ $N_0\approx 1.72\times 10^{17}$

3. Atoms decayed in three half-lives

After $3$ half-lives, fraction remaining is $\left(\frac12\right)^3=\frac18$

So fraction decayed is $1-\frac18=\frac78$

Hence number of uranium atoms that have decayed is $N_{\text{decayed}}=\frac78N_0$

$N_{\text{decayed}}=\frac78\times 1.72\times 10^{17}$ $N_{\text{decayed}}\approx 1.505\times 10^{17}$

4. Total number of $\alpha$ particles emitted

Each decayed uranium nucleus emits $8\alpha$ particles in reaching lead. Therefore, $N_\alpha=8\times N_{\text{decayed}}$

$N_\alpha=8\times 1.505\times 10^{17}$ $N_\alpha\approx 1.204\times 10^{18}$

Given $N_\alpha=Z\times 10^{18}$ we get $Z\approx 1.2$

5. Final answer

$\boxed{Z=1.2}$