Answer
$\frac{1}{2}$
Work Step by Step
We know that $\gamma=\dfrac{C_{\rm P}}{C_{\rm V}}$ and that $C_{\rm P}=R+C_{\rm V}$
Hence,
$$\gamma=\dfrac{R+C_{\rm V}}{C_{\rm V}}$$
$$1.5=\dfrac{R+C_{\rm V}}{C_{\rm V}}$$
$$1.5C_{\rm V}= R+C_{\rm V} $$
Hence,
$$ C_{\rm V}= 2R \tag 1$$
Let's assume that $n_1$ is the moles of the monatomic gas and $n_2$ is the moles of the diatomic gas.
We know that the thermal energy of the system is given by
$$E_{th}=E_{th,1}+E_{th,2}$$
where $E_{th,1}$ is for the monatomic gas and is given by $3nRT/2$ while $E_{th,2}$ is for the diatomic gas and is given by $5nRT/2$
$$E_{th}=\dfrac{3n_1RT}{2}+\dfrac{5n_2RT}{2}$$
Recalling that $E_{th}=n C_{\rm V}T$ where $n$ here is the total moles of the system.
$$n C_{\rm V} \color{red}{\bf\not} T=\dfrac{3n_1R \color{red}{\bf\not} T}{2}+\dfrac{5n_2R \color{red}{\bf\not} T}{2}$$
$$2n C_{\rm V} = 3n_1R + 5n_2R $$
Hence,
$$2(n_1+n_2) C_{\rm V} =(3n_1 + 5n_2)R $$
Plugging from (1);
$$2(n_1+n_2) 2 \color{red}{\bf\not} R=(3n_1 + 5n_2) \color{red}{\bf\not} R $$
$$4(n_1+n_2) =(3n_1 + 5n_2) $$
$$4n_1+4n_2 =(3n_1 + 5n_2) $$
Therefore,
$$\boxed{n_1=n_2}$$
Therefore, the fraction of the molecules needed to be monatomic is half the molecules.
