For monatomic molecule, degree of freedom =3.
The formula for the average kinetic energy is given by
Kavg=21fKBT...(1)
From equation (1), it follows that
0.414eV×1eV1.6×10−19J=23×1.38×10−23Jmol−1K−1×T⇒T=3×1.38×10−23Jmol−1K−10.414eV×1eV1.6×10−19J×2=3200K