A dye fades as it reacts with hydroxide ions…

Step 1. The pseudo first order constants. $k' = \dfrac{0.693}{t_{1/2}}$. The three runs give $0.0220\ \text{s}^{-1}$, $0.0439\ \text{s}^{-1}$ and $0.0877\ \text{s}^{-1}$. For the run at $0.080\ \text{mol/L}$, $k' = \dfrac{0.693}{15.8\ \text{s}} = 0.0439\ \text{s}^{-1}$.

Step 2. The test. Each time the hydroxide concentration doubles, $k'$ doubles. So $k'$ is proportional to $[\text{OH}^-]$, and the reaction is pseudo first order.

Step 3. The true rate constant. $k = \dfrac{k'}{[\text{OH}^-]} = \dfrac{0.0439\ \text{s}^{-1}}{0.080\ \text{mol/L}} = 0.549\ \text{L mol}^{-1}\,\text{s}^{-1}$. The other two runs give almost the same value.

Step 4. Ninety percent. $t = \dfrac{2.303}{k'}\log 10 = \dfrac{2.303}{0.0439\ \text{s}^{-1}} = 52.5\ \text{s}$.