For an elementary chemical reaction, the Arrhenius plot is given...

According to the Arrhenius equation:

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

For a plot of $\ln k$ versus $1/T$, the y-intercept corresponds to $\ln A$. From the given graph, the line intersects the y-axis at $6$.

Therefore, $\ln A = 6$.

We are given:
$E_a = 6.64 \text{ kJ mol}^{-1} = 6640 \text{ J mol}^{-1}$
$R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$
$k = e^2 \text{ min}^{-1}$, which gives $\ln k = \ln(e^2) = 2$

Substituting these values into the Arrhenius equation:

$2 = 6 - \dfrac{6640}{8.3 \times T}$

$\dfrac{6640}{8.3 \times T} = 6 - 2$

$\dfrac{6640}{8.3 \times T} = 4$

$T = \dfrac{6640}{4 \times 8.3}$

$T = \dfrac{6640}{33.2}$

$T = 200 \text{ K}$

Answer: $200$ K