This solution uses the base-10 (logarithmic) form of the Arrhenius equation, which is often easier for board arithmetic.
Step 1: The logarithmic Arrhenius relation between two temperatures is
\[ \log\frac{k_2}{k_1} = \frac{E_a}{2.303\,R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \]
Step 2: Insert the data with \(k_2/k_1 = 3.5\), so \(\log 3.5 = 0.5441\) and \(\frac{1}{500}-\frac{1}{700} = 5.714\times10^{-4}\ \text{K}^{-1}\).
Step 3: Rearranging for the activation energy,
\[ E_a = \frac{2.303 \times R \times \log(k_2/k_1)}{\left(\frac{1}{T_1}-\frac{1}{T_2}\right)} = \frac{2.303 \times 8.314 \times 0.5441}{5.714\times10^{-4}} \]
\[ = \frac{10.42}{5.714\times10^{-4}} \approx 18230\ \text{J mol}^{-1} = 18.23\ \text{kJ mol}^{-1} \]
\[ \boxed{E_a \approx 18.23\ \text{kJ mol}^{-1}} \]
Step 4: For the pre-exponential factor use \(\log k = \log A - \dfrac{E_a}{2.303\,R\,T}\), hence \(\log A = \log k_1 + \dfrac{E_a}{2.303\,R\,T_1}\).
Step 5: Compute the second term: \(2.303\times 8.314\times 500 = 9575\), so \(\dfrac{18230}{9575} = 1.904\). With \(\log(0.02) = -1.699\),
\[ \log A = -1.699 + 1.904 = 0.205 \implies A = 10^{0.205} \approx 1.60\ \text{s}^{-1} \]
\[ \boxed{A \approx 1.60\ \text{s}^{-1}} \]
The units of A match those of k (s\(^{-1}\)), as expected for a first-order rate constant.