Prove that a 10-Degree Temperature Increase Doubles the Rate Constant (k), when the Activation Energy is Approximately 50 kJ/mol
Your method and your mathematics seem perfectly fine, and your calculated result is correct. You are also quite correct that the result will change depending on the value of $E_\mathrm a$ that you choose.

The problem you're running into is the assumption that the stated rule of thumb holds exactly, regardless of temperature. It doesn't.



I would set up the problem as follows. Consider the ratio $\rho$ between the reaction occurring at some Kelvin temperature $T$, and at $T+10$:

$$
\rho = {e^{-E_\mathrm a\over R\,\left(T+10\right)}\over
        e^{-E_\mathrm a\over RT}}
     = e^{{E_\mathrm a\over R}\left({1\over T} - {1\over T+10}\right)}
     = e^{{E_\mathrm a\over R}\cdot {10\over{T^2+10T}}}
$$

Plotting $\rho$ as a function of $T$ for $E_\mathrm a = 50~\mathrm{kJ\over mol}$, it becomes immediately clear that the rule of thumb does not hold over a very wide temperature range:



In fact, it only holds strictly for one specific temperature, which is why your math led you to a single temperature as your answer. By my calculations, using a value of $8.3144598~\mathrm{J\over mol\, K}$ for $R$, that temperature is $289.56~K$, which is equal to your $T_1=290~K$ to within three significant figures.

The gray box in the figure marks where the rule holds to within $20\%$ – that is, where $\rho$ falls between $1.6$ and $2.4$. This corresponds to a temperature range of $257~\mathrm K$ to $353~\mathrm K$.