Answer
$2.86\times10^{9}\,y$
Work Step by Step
Mass of $^{238}U$ decayed=3.22 mg $^{206}Pb\times\frac{238\,mg\,\,^{238}U}{206\,mg\,\,^{206}Pb}=3.72\,mg$
So, the original amount $A_{0}=6.73\,mg+3.72\,mg=10.45\,mg$
and the amount left is $A=6.73\,mg$
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{4.51\times10^{9}\,y}=1.5366\times10^{-10}\,y^{-1}$
Recall that $\ln(\frac{A_{0}}{A})=kt$, where $t$ is the age.
$\implies \ln(\frac{10.45}{6.73})=0.440=1.5366\times10^{-10}\,y^{-1}(t)$
Or $t=\frac{0.440}{1.5366\times10^{-10}\,y^{-1}}=2.86\times10^{9}\,y$
