To solve this problem, we need to determine the time required for the completion of $99\%$ of a first-order reaction.$\newline$For a first-order reaction, the integrated rate equation is given by:$\newline \ln \left( \frac{[A]_0}{[A]} \right) = kt \newline$where $[A]_0$ is the initial concentration and $[A]$ is the concentration at time $t. \newline$For $99\%$ completion, $[A] = 0.01[A]_0. \newline$Substitute this into the equation:$\newline \ln \left( \frac{[A]_0}{0.01[A]_0} \right) = kt \newline \ln \left( \frac{1}{0.01} \right) = kt \newline \ln(100) = kt \newline$Since $\ln(100) = 4.606,$ we have:$\newline 4.606 = kt \newline$Solving for $t,$ we get:$\newline t = \frac{4.606}{k} \newline$Therefore, the time required for the completion of $99\%$ of the reaction is given by:$\newline t = \frac{4.606}{k} \newline$This corresponds to Option 3.