Energy possessed by the ideal gas at $27^{\circ}\,C$ is

$E _1=3\left(\frac{3}{2} R \times 300\right)=\frac{2700 R }{2}$

Energy possessed by the ideal gas at $227^{\circ} C$ is

$E _2=2\left(\frac{3 R }{2} \times 500\right)=1500\,R$

If $T$ be the equilibrium temperature, of the mixture, then its energy will be $E _{ m }=5\left(\frac{3 RT }{2}\right)$

Since, energy remains conserved,

$E _{ m }= E _1+ E _2$

or $5\left(\frac{3 R T}{2}\right)=\frac{2700 R }{2}+1500 R$ or $T =380 K$ or $107^{\circ}\,C$