The half-life of $$Au^{198}$$ is 2.7 days. The activity of 1.50 mg of $$Au^{198}$$, if its atomic weight is 198 g mol$$^{-1}$$ is, ($$N_A = 6 \times 10^{23}$$ mol$$^{-1}$$)

The activity of a radioactive sample is $$A = \lambda N$$, where $$\lambda = \frac{\ln 2}{t_{1/2}}$$ is the decay constant and $$N$$ is the number of atoms.

The number of atoms in $$1.50$$ mg of $$\text{Au}^{198}$$ is $$N = \frac{m}{M} \times N_A = \frac{1.50 \times 10^{-3}}{198} \times 6 \times 10^{23} = \frac{1.50 \times 6 \times 10^{20}}{198} = \frac{9 \times 10^{20}}{198} = 4.545 \times 10^{18}$$.

The decay constant is $$\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{2.7 \times 86400 \text{ s}} = \frac{0.693}{233280} = 2.971 \times 10^{-6}$$ s$$^{-1}$$.

The activity is $$A = \lambda N = 2.971 \times 10^{-6} \times 4.545 \times 10^{18} = 1.350 \times 10^{13}$$ disintegrations per second (Bq).

Converting to Curie (1 Ci = $$3.7 \times 10^{10}$$ Bq): $$A = \frac{1.350 \times 10^{13}}{3.7 \times 10^{10}} \approx 365$$ Ci. Given the approximations used, this is closest to $$357$$ Ci.