1. Use Arrhenius equation in ratio form

For a reaction, $k = A e^{-E_a/(RT)}$

Taking ratio at two temperatures $T_1$ and $T_2$, $\ln\left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

Since rate is proportional to rate constant, we can use the same relation for rates.

2. Given data

Temperature decreases from $40^\circ C$ to $30^\circ C$. Thus, $T_1 = 40^\circ C = 313\,K$ $T_2 = 30^\circ C = 303\,K$

The rate decreased by $3.555$ times, so $\frac{k_1}{k_2} = 3.555$ where $k_1$ is at $313\,K$ and $k_2$ is at $303\,K$.

Hence, $\ln\left(\frac{k_1}{k_2}\right)=\ln(3.555)=1.268$

So we write, $\ln\left(\frac{k_{313}}{k_{303}}\right)=\frac{E_a}{R}\left(\frac{1}{303}-\frac{1}{313}\right)$

3. Substitute values

$1.268 = \frac{E_a}{8.314}\left(\frac{1}{303}-\frac{1}{313}\right)$

Now, $\frac{1}{303}-\frac{1}{313} = \frac{313-303}{303\times 313} = \frac{10}{94839}$

Thus, $1.268 = \frac{E_a}{8.314}\cdot \frac{10}{94839}$

4. Solve for $E_a$

$E_a = 1.268 \times 8.314 \times \frac{94839}{10}$

$E_a \approx 1.268 \times 8.314 \times 9483.9$

First, $1.268 \times 8.314 \approx 10.54$

Then, $E_a \approx 10.54 \times 9483.9 \approx 99958\,\text{J mol}^{-1}$

$E_a \approx 1.00\times 10^5\,\text{J mol}^{-1}$

Converting to kJ mol$^{-1}$, $E_a \approx 100\,\text{kJ mol}^{-1}$

5. Final answer

$\boxed{100}$

6. Comparison with stored answer

Stored correct answer = $100$

Our derived answer matches the stored answer.