Step 1: Write the integrated rate law for a first-order reaction.
\[ k = \frac{2.303}{t} \log\!\left(\frac{[A]_0}{[A]_t}\right) \] where $[A]_0$ is the initial concentration, $[A]_t$ is the concentration at time $t$.
Step 2: Identify the given values.
$[A]_0 = 0.6\ \text{mol L}^{-1}$, $[A]_t = 0.2\ \text{mol L}^{-1}$, $t = 5\ \text{min}$.
Step 3: Calculate the concentration ratio.
\[ \frac{[A]_0}{[A]_t} = \frac{0.6}{0.2} = 3 \]
Step 4: Take the logarithm.
We are given $\log 3 = 0.4771$.
Step 5: Substitute all values.
\[ k = \frac{2.303}{5} \times 0.4771 = 0.4606 \times 0.4771 \approx 0.2197\ \text{min}^{-1} \]
Step 6: Round to correct significant figures.
\[ k \approx 0.219\ \text{min}^{-1} \] The units $\text{min}^{-1}$ confirm this is a first-order rate constant.
Step 7: State the final answer.
\[ \boxed{k \approx 0.219\ \text{min}^{-1}} \]