Nobody hands you $A$. So measure the reaction twice, at two temperatures, and divide one Arrhenius equation by the other: $A$ cancels and never comes back: $\ln\dfrac{k_2}{k_1} = \dfrac{E_a}{R}\left(\dfrac{1}{T_1} - \dfrac{1}{T_2}\right)$. The subscripts are a pairing, so state the convention once and hold it: subscript 1 is the first (cooler) run and subscript 2 the second (warmer) one: $k_1$ belongs with $T_1$, $k_2$ with $T_2$, both temperatures in kelvin, and $E_a$ comes out in joules per mole because $R$ is in joules. The unknown here is $E_a$.

Watch the reciprocals. They arrive as $\dfrac{1}{T_1} - \dfrac{1}{T_2}$, and with $T_2$ the hotter run that difference is POSITIVE — the smaller number is being subtracted. Swap them and $E_a$ comes out negative, which is a hill that runs downhill: not a thing. Let the units keep you honest one way only — if they refuse to cancel into J/mol the rearrangement is wrong, no appeal; units that do work out never prove you right.

Measure at five temperatures instead of two and you can plot $\ln k$ against $1/T$. Taking $\ln$ of Arrhenius turns it into a straight line, $\ln k = \ln A - \dfrac{E_a}{R}\cdot\dfrac{1}{T}$, whose slope (in kelvin, and always negative) is $-E_a/R$. So $E_a = -R \times \text{slope}$ — the minus sign in the formula is exactly what turns a downhill line into an uphill barrier. Same physics, two roads, and the graph uses every point you measured instead of just two.