Given, Half-life of radium, $T_{1/2} = 1500 \text{ y}$ Initial mass, $N_0 = 1 \text{ g} = 1000 \text{ mg}$ Final mass, $N = N_0 - 1 \text{ mg} = 1000 \text{ mg} - 1 \text{ mg} = 999 \text{ mg}$ The decay constant $\lambda$ is given by, $$ \begin{aligned} \lambda &= \frac{\ln 2}{T_{1/2}} \\ &= \frac{0.6931}{1500} \text{ y}^{-1} \\ &\approx 4.621 \times 10^{-4} \text{ y}^{-1} \end{aligned} $$ Using the radioactive decay law, $N = N_0 e^{-\lambda t}$, we have: $$ \begin{aligned} \frac{N}{N_0} &= e^{-\lambda t} \\ \ln\left(\frac{N}{N_0}\right) &= -\lambda t \\ t &= -\frac{1}{\lambda} \ln\left(\frac{N}{N_0}\right) \end{aligned} $$ Substituting the values: $$ \begin{aligned} t &= -\frac{1}{4.621 \times 10^{-4}} \ln\left(\frac{999}{1000}\right) \\ &= -\frac{1}{4.621 \times 10^{-4}} \ln(0.999) \\ &= -\frac{1}{4.621 \times 10^{-4}} \times (-0.0010005) \\ &\approx 2.165 \text{ y} \end{aligned} $$ Thus, the time required is approximately $2.165 \text{ years}$.