We know the relation between absolute temperature and average speed of the molecule inside the container is, 

$v_{avg}\propto \sqrt{T}$, it means, 

$$\begin{gathered} \frac{{\left(v_{avg}\right)}_1}{{\left(v_{avg}\right)}_2}=\frac{\sqrt{T_1}}{\sqrt{T_2}} \\ \Rightarrow \frac{{\left(v_{avg}\right)}_1}{4{\left(v_{avg}\right)}_1}=\frac{\sqrt{300}}{\sqrt{T_2}} \\ \Rightarrow T_2=4800 \mathrm{K} \end{gathered}$$

Now use the relation between centigrade scale and Celsius scale, 

$\Rightarrow T_2=4800-273=4527°\mathrm{C}$.