Let the initial moles of U-238 be $N_0$ and the current moles be $N$. Then moles of Pb-206 produced $=N_0-N$.

The mass ratio $\frac{m(\text{Pb-206})}{m(\text{U-238})}=\frac{206(N_0-N)}{238N}=7.$

Solving: $206(N_0-N)=7\cdot 238N\Rightarrow 206N_0=(206+1666)N=1872N.$

$\frac{N_0}{N}=\frac{1872}{206}\approx \mathrm{9.087.}$

Using first-order decay $N=N_0{e}^{-\lambda t}$ with $\lambda =\frac{\ln 2}{t_{1/2}}$:

$t=\frac{t_{1/2}}{\ln 2}\ln \frac{N_0}{N}=\frac{4.5\times 10^9}{0.693}\cdot \ln \mathrm{9.087.}$

$\ln 9.087\approx \mathrm{2.207.}$

$t\approx \frac{4.5\times 10^9\cdot 2.207}{0.693}\approx 1.433\times 10^{10}\text{years}=143.3\times 10^8\text{years}.$

$\therefore P\approx 143.$