A first order reaction has k = 0.0420 min⁻¹ and starts at 0.36 mol/L…

Step 1. The exponent. $kt = 0.0420\ \text{min}^{-1} \times 25\ \text{min} = 1.05$.

Step 2. The share left. $\dfrac{[R]}{[R]_0} = e^{-1.05} = 0.350$, which is $35.0\ \%$.

Step 3. The concentration. $[R] = 0.36\ \text{mol/L} \times 0.350 = 0.126\ \text{mol/L}$.

Step 4. For 90 %. If 90 % has reacted, 10 % is left, so $\dfrac{[R]_0}{[R]} = 10$. Then $t = \dfrac{2.303}{k}\log 10 = \dfrac{2.303}{0.0420\ \text{min}^{-1}} = 54.8\ \text{min}$.