$T_{2}=?$

That the Final $rms$ speed will be Double of inital.

$\frac{V_{rms_{2}}}{V_{rms_{1}}}=\sqrt{\frac{T_{2}}{T_{1}}}=\sqrt{\frac{T_{2}}{273}}$

$V_{rms_{2}} =2 V_{rms_{1}}$

$\Rightarrow \quad 2 =\sqrt{\frac{T_{2}}{273}}$

$\Rightarrow \quad T_{2} =4 \times 273=1092 \; K.$

$\quad T_{2} =819^{\circ} C$

The temperature raised $=819^{\circ} C$