Given here, $\frac{m_{\mathrm{Ar}}}{m_{\mathrm{O}}}=\frac{3}{2}$.

As argon is monoatomic gas, so it has three degree of freedom $\left(f\right)$ and oxygen is a diatomic gas, so its degree of freedom is five.

We know that average kinetic energy per molecule is given by $KE=\frac{f}{2}kT$, where, $k$ is Boltzmann's constant and $T$ is temperature.

Here, $T=27°\mathrm{C}$ is fixed for both gas.

So, $\frac{{\left(KE\right)}_{\mathrm{Ar}}}{{\left(KE\right)}_{\mathrm{O}}}=\frac{{\displaystyle \frac{3}{2}}kT}{{\displaystyle \frac{5}{2}}kT}=\frac{3}{5}$