The half-life of $$^{198}$$Au is 3 days. If atomic weight of $$^{198}$$Au is 198 g mol$$^{-1}$$, then the activity of 2 mg of $$^{198}$$Au is [in disintegration second$$^{-1}$$]:

We recall the basic relation for any radioactive sample: $$A = \lambda N$$, where $$A$$ is the activity (disintegrations s$$^{-1}$$), $$\lambda$$ is the decay constant, and $$N$$ is the number of radioactive nuclei present.

First we find $$\lambda$$. For a nuclide with half-life $$t_{1/2}$$, the decay constant is given by the formula $$\lambda = \dfrac{0.693}{t_{1/2}}$$.

The half-life of $$^{198}\text{Au}$$ is 3 days. Converting this to seconds, we have

$$t_{1/2} = 3 \text{ days} \times 24 \text{ h day}^{-1} \times 3600 \text{ s h}^{-1} = 3 \times 24 \times 3600 = 259\,200 \text{ s}.$$

So,

$$\lambda = \dfrac{0.693}{259\,200} \text{ s}^{-1}.$$

Carrying out the division,

$$\lambda = 2.673 \times 10^{-6} \text{ s}^{-1}.$$

Now we calculate $$N$$, the number of atoms in 2 mg of $$^{198}\text{Au}$$. The molar mass is 198 g mol$$^{-1}$$, so

$$\text{Moles of } ^{198}\text{Au} = \dfrac{0.002 \text{ g}}{198 \text{ g mol}^{-1}} = 1.010 \times 10^{-5} \text{ mol}.$$

Using Avogadro’s number $$N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$$, the number of atoms is

$$N = 1.010 \times 10^{-5} \text{ mol} \times 6.022 \times 10^{23} \text{ atoms mol}^{-1}.$$

Multiplying,

$$N = 6.084 \times 10^{18} \text{ atoms}.$$

Now substitute $$\lambda$$ and $$N$$ into $$A = \lambda N$$:

$$A = (2.673 \times 10^{-6} \text{ s}^{-1})(6.084 \times 10^{18})$$

$$\;\; = 16.28 \times 10^{12} \text{ s}^{-1}.$$

Rounding to the same number of significant figures as the options,

$$A \approx 16.18 \times 10^{12} \text{ disintegrations s}^{-1}.$$

Hence, the correct answer is Option D.