1. Use the Arrhenius equation for a first-order reaction:

$k = A e^{-E_a/(RT)}$

Taking natural logarithm:

$\ln k = \ln A - \frac{E_a}{RT}$

So,

$\frac{E_a}{RT} = \ln\left(\frac{A}{k}\right)$

2. Substitute the given values:

$A = 4 \times 10^{13}\ \text{s}^{-1}, \qquad k = 2 \times 10^{-6}\ \text{s}^{-1}$

Hence,

$\frac{A}{k} = \frac{4 \times 10^{13}}{2 \times 10^{-6}} = 2 \times 10^{19}$

Thus,

$\ln\left(\frac{A}{k}\right) = \ln(2 \times 10^{19}) = \ln 2 + 19\ln 10$

Given:

$\ln 10 = 2.3, \qquad \log 2 = 0.30$

Since $\ln 2 = 2.3 \times \log 2 = 2.3 \times 0.30 = 0.69$,

$\ln\left(\frac{A}{k}\right) = 0.69 + 19(2.3) = 0.69 + 43.7 = 44.39$

3. Now use:

$\frac{E_a}{RT} = 44.39$

Given:

$E_a = 8.3\ \text{kJ mol}^{-1} = 8300\ \text{J mol}^{-1}, \qquad R = 8.3\ \text{J K}^{-1}\text{mol}^{-1}$

So,

$\frac{8300}{8.3\,T} = 44.39$

$\frac{1000}{T} = 44.39$

$T = \frac{1000}{44.39} \approx 22.5\ \text{K}$

4. The question asks for the value in the form:

$\underline{\hspace{2cm}} \times 10^{-1}\ \text{K}$

Since

$22.5\ \text{K} = 225 \times 10^{-1}\ \text{K}$

Therefore, the required integer is:

$\boxed{225}$

5. Comparison with stored correct answer:

Stored correct answer = $225$.

Our derived answer matches it.