To a first approximation, the activation energy of a single given reaction is a constant at all temperatures (see reasons here, although more precisely, there is some temperature variation). However, the activation energies for both reactions need not be the same.
If you consider both reactions at $\pu{321 K}$,
$$\begin{align}
\ln k_1 &= \ln k_2 \tag{1}\\
\ln A_1 - \frac{E_{\mathrm A}^{(1)}}{(\pu{321 K})R} &= \ln A_2 - \frac{E_\mathrm{A}^{(2)}}{(\pu{321 K})R} \tag{2}\\[4pt]
\ln A_1 - \ln A_2 &= \frac{E_{\mathrm A}^{(1)} - E_{\mathrm A}^{(2)}}{(\pu{321 K})R} \tag{3} \\[4pt]
&= \frac{\pu{71.9 kJ mol-1} - \pu{142.8 kJ mol-1}}{(\pu{321 K})(\pu{8.31446 J K-1 mol-1})} \tag{3} \\[4pt]
&= -26.56 \tag{4}
\end{align}$$
Now, denote the desired temperature at which $k_1 = 2k_2$ as $T^*$. At this new temperature, we have
$$\begin{align}
\ln k_1 &= \ln 2 + \ln k_2 \tag{5}\\
\ln A_1 - \frac{E_{\mathrm A}^{(1)}}{T^*R} &= \ln 2 + \ln A_2 - \frac{E_\mathrm{A}^{(2)}}{T^*R} \tag{6}\\[4pt]
\ln A_1 - \ln A_2 - \ln 2 &= \frac{E_{\mathrm A}^{(1)} - E_{\mathrm A}^{(2)}}{T^*R} \tag{7}
\end{align}$$
Here we can reuse the previously calculated value for $\ln A_1 - \ln A_2$, and we can also plug in the same activation energies:
$$\begin{align}
T^* &= \frac{E_{\mathrm A}^{(1)} - E_{\mathrm A}^{(2)}}{R(\ln A_1 - \ln A_2 - \ln 2)} \tag{8} \\[4pt]
&= \frac{\pu{71.9 kJ mol-1} - \pu{142.8 kJ mol-1}}{(\pu{8.31446 J K-1 mol-1})(-26.56 - 0.693)} \tag{9} \\[4pt]
&= \pu{313 K}. \tag{10}
\end{align}$$
