To determine how long it takes for the concentration of a reactant to decrease from 7.2 mol L$^{-1}$ to 0.9 mol L$^{-1}$, we use the formula for the first-order reaction:

$ t = \frac{2.303}{k} \log \frac{a}{a-x} $

Where:

$ k = 0.03 \, \text{s}^{-1} $ is the rate constant.

$ a = 7.2 \, \text{mol L}^{-1} $ is the initial concentration.

$ a-x = 0.9 \, \text{mol L}^{-1} $ is the final concentration.

Plug these values into the equation:

$ \begin{aligned} t & = \frac{2.303}{0.03} \log \frac{7.2}{0.9} \\ & = \frac{2.303}{0.03} \log 8 \\ & = \frac{2.303}{0.03} \times 3 \times \log 2 \\ & = \frac{2.303}{0.03} \times 3 \times 0.301 \\ & = 69.3 \, \text{s} \end{aligned} $

Therefore, it takes 69.3 seconds for the concentration to decrease from 7.2 mol L$^{-1}$ to 0.9 mol L$^{-1}$.