For a first-order reaction, the integrated rate equation is given by: $t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$ where: $t$ is the time required, $k$ is the rate constant = $4.606 \times 10^{-3} \text{ s}^{-1}$, $[R]_0$ is the initial concentration (or mass) = 2.0 g, $[R]$ is the final concentration (or mass) = 0.2 g.

Substituting the values into the equation: $t = \frac{2.303}{4.606 \times 10^{-3}} \log \left( \frac{2.0}{0.2} \right)$ $t = \frac{1}{2 \times 10^{-3}} \log(10)$ Since $\log(10) = 1$: $t = \frac{1000}{2} = 500 \text{ s}$ .