They call it chemical kinetics?

Firstly, we need to determine what the values of $A$ and $E_A$ are. We can do this by estimating two points on the given graph - I'll choose $(0.75, 0.25)$ and $(1.5, 0.5)$ (I have omitted units to make the calculations more readable). Since $A$ and $E_A$ are constant regardless of our choice of $T$ , we can write:

$\displaystyle \frac{0.25}{e^{-\frac{E_{A}}{0.75\text{K} * R}}}=\frac{0.5}{e^{-\frac{E_{A}}{1.5 * R}}}$

Which, through simplification, gives us:

$\displaystyle e^{\displaystyle\frac{E_{A}}{R}\displaystyle\left(\frac{1}{0.75}-\frac{1}{1.5}\right)}=2$

Meaning that $E_{A}=\frac{3R\ln\left(2\right)}{2}$ and $A = 0.25e^{\frac{E_{A}}{0.75R}} = 1$

Now we need to determine the formula for the inflection point of $k(T)$ . We have $k\left(T\right)=Ae^{-\frac{E_{A}}{RT}}$ meaning that $k'(T) = \frac{E_{A}}{RT^{2}}Ae^{-\frac{E_{A}}{RT}}$ and $k''(T) = -\frac{2E_{A}}{RT^{3}}Ae^{-\frac{E_{A}}{RT}}+\frac{E_{A}^{2}}{R^{2}T^{4}}Ae^{-\frac{E_{A}}{RT}}$ We can vastly simplify $k''(T)$ by factoring out $\frac{E_{A}}{R}Ae^{-\frac{E_{A}}{RT}}$ , giving us $k''(T) = \frac{E_{A}}{R}Ae^{-\frac{E_{A}}{RT}}\left(-\frac{2}{T^{3}}+\frac{E_{A}}{RT^{4}}\right)$ . To find the inflection point of $k(T)$ , we need to find a $T$ such that $k''(T) = 0$ . This can only happen if $-\frac{2}{T^{3}}+\frac{E_{A}}{RT^{4}}=0$ and since the $T$ we want will be positive, we can multiply this expression by $T^3$ to give us $-2+\frac{E_{A}}{RT}=0$ . There is therefore an inflection point at $T = \frac{E_{A}}{2R} = \frac{3}{4}\ln\left(2\right) \text{K}$

The value of $k$ at this inflection point is $k' = Ae^{-\frac{E_{A}}{RT}} = Ae^{-2}$ . Finally, we conclude that $|X| = \boxed{2}$