The kinetic energy is given by $KE=\frac{f}{2}kT$.

Let the kinetic energy be $K_1$ at $27 °\mathrm{C}$ and the kinetic energy be $K\text{'}$ at $T\text{'}$ temperature. As given in the question, $K\text{'}=2K_1$. Hence, 

$$\begin{gathered} \frac{K\text{'}}{K_1}=\frac{T\text{'}}{T} \\ \Rightarrow 2=\frac{T\text{'}}{300} \\ \Rightarrow T\text{'}=600 \mathrm{K} \end{gathered}$$

In Celsius the temperature is $T\text{'}-273=327 °\mathrm{C}$.