The eclipsed form of ethane lies 12 kJ/mol above the staggered form…

Step 1. At $298\ \text{K}$. $\dfrac{E}{RT} = \dfrac{12000\ \text{J/mol}}{8.314\ \text{J/(mol K)} \times 298\ \text{K}} = 4.843$. So $e^{-4.843} = 0.00788$, which is $0.788\ \%$.

Step 2. At $350\ \text{K}$. $\dfrac{12000}{8.314 \times 350} = 4.124$. So $e^{-4.124} = 0.0162$, which is $1.62\ \%$.