A reaction has an activation energy of 61.0 kJ/mol…

Step 1. The temperature part. $\dfrac{1}{305\ \text{K}} - \dfrac{1}{325\ \text{K}} = 2.018 \times 10^{-4}\ \text{K}^{-1}$.

Step 2. The logarithm of the ratio. $\ln\dfrac{k_2}{k_1} = \dfrac{61\,000}{8.314}\ \text{K} \times 2.018 \times 10^{-4}\ \text{K}^{-1} = 7337 \times 2.018 \times 10^{-4} = 1.480$.

Step 3. The ratio. $\dfrac{k_2}{k_1} = e^{1.480} = 4.39$.

Step 4. The new rate constant. $k_2 = 2.6 \times 10^{-3}\ \text{s}^{-1} \times 4.39 = 0.0114\ \text{s}^{-1}$.