Answer
$2.7\times10^{9}\,yr$
Work Step by Step
Mass in grams of U-238 that is required to form 0.438 g of Pb-206
$=0.438\,g\,Pb\mbox{-}206\times\frac{1\,mol\,Pb\mbox{-}206}{206\,g\,Pb\mbox{-}206}\times\frac{1\,mol\,U\mbox{-}238}{1\,mol\,Pb\mbox{-}206}\times\frac{238\,g\,U\mbox{-}238}{1\,mol\,U\mbox{-}238}$
$=0.506\,g$
$k=\frac{0.693}{t_{1/2}}=\frac{0.693}{4.5\times10^{9}\,yr}=1.54\times10^{-10}/yr$
Recall that $\ln(\frac{N}{N_{0}})=-kt$
$\implies t=-\frac{\ln(\frac{N}{N_{0}})}{k}=\frac{\ln(\frac{1.00\,g}{1.00\,g+0.506\,g})}{1.54\times10^{-10}/yr}=2.7\times10^{9}\,yr$
