\(A_{0}=16\) dis \(\min ^{-1} g^{-1}\)

\(A=12\) dis \(\min ^{-1} g^{-1}\)

\(\mathrm{t}_{1 / 2}=5760\) years

Now, \(\lambda  = \frac{{0.693}}{{{t_{1/2}}}}\)

\(\lambda=\frac{0.693}{5760}\) per year

Then, from, \(t=\frac{2.303}{\lambda} \log _{10} \frac{\mathrm{A}_{0}}{\mathrm{A}}\)

\(=\frac{2.303 \times 5760}{0.693} \log _{10} \frac{16}{12}\)

\( = \frac{{2.303 \times 5760}}{{0.693}}{\log _{10}}1.333\)

\( = \frac{{2.303 \times 5760 \times 0.1249}}{{0.693}}\)

\(=2390.81 \approx 2391\) years