For a first-order reaction, the integrated rate equation is given by: $k = \frac{2.303}{t} \log \frac{[R]_0}{[R]}$ Given: The substance decomposes to $40\%$, meaning the remaining concentration is $[R] = 40\%$ of the initial concentration $[R]_0$. So, $\frac{[R]_0}{[R]} = \frac{100}{40} = 2.5$ Time, $t = 560 \text{ s}$ Substituting the values into the rate equation: $k = \frac{2.303}{560} \log(2.5)$ Given $\log 2.5 = 0.3979$, $k = \frac{2.303 \times 0.3979}{560} \text{ s}^{-1} = \frac{0.91636}{560} \text{ s}^{-1} = 1.636 \times 10^{-3} \text{ s}^{-1}$ To find the rate constant in $\text{min}^{-1}$, we multiply by $60 \text{ s/min}$: $k = 1.636 \times 10^{-3} \text{ s}^{-1} \times 60 \text{ s/min} = 0.09816 \text{ min}^{-1} = 9.816 \times 10^{-2} \text{ min}^{-1}$ Since none of the options match the correct calculated value, 'None of the above' is the correct choice.