Given. Half-life of $\mathrm{C}^{4}, t_{1 / 2}=5760$ years Initial concentration of sample of $\mathrm{C}^{4}, \mathrm{N}_{0}=200 \mathrm{mg}$ Final concentration of sample of $\mathrm{C}^{4}, \mathrm{N}_{1}=2 \mathrm{S} \mathrm{mg}$ The given decay is radioactive and all radiouctive decay follows first order kinetics.
$$
t_{1 / 2}=\frac{0.693}{\lambda}
$$
$\therefore$
$$
\lambda=\frac{0.693}{t_{1 / 2}}=\frac{0.693}{5760} \mathrm{yr}^{-1}
$$
We know that, for first order reaction
$\begin{aligned} t &=\frac{2.303}{\lambda} \log \frac{N_{0}}{N_{1}}=\frac{2303}{0.693} \log \frac{[200]}{125} \\ &=\frac{2303 \times 5760}{0.693} \log 8 \\ &=17,28678 \mathrm{yrs} \approx 17.280 \mathrm{yrs} \end{aligned}$
Therefore, the time taken to change 200 me to $25 \mathrm{mg}$ is $17,286.78 \mathrm{yr}$, which is very close to option (d)