For a first-order reaction, the time required is given by the integrated rate equation:

$t = \frac{2.303}{k} \log \frac{[R]_0}{[R]}$

Given data: Rate constant, $k = 10^{-2} \text{ s}^{-1}$ Initial amount, $[R]_0 = 20 \text{ g}$ Final amount, $[R] = 5 \text{ g}$

Substituting the values into the equation: $t = \frac{2.303}{10^{-2}} \log \left( \frac{20}{5} \right)$ $t = 2.303 \times 10^2 \times \log(4)$ $t = 230.3 \times 2 \log(2)$ $t = 230.3 \times 2 \times 0.3010$ $t = 230.3 \times 0.6020 \approx 138.64 \text{ s}$

Therefore, the time required is approximately $138.6 \text{ s}$.