For mixture of gases, we have
$\frac{\left(n_{1}+n_{2}\right)}{(\gamma-1)}=\frac{n_{1}}{\left(\gamma_{1}-1\right)}+\frac{n_{2}}{\left(\gamma_{2}-1\right)}$ ...(1)
Given: $n_{1}=2$ (for helium); $\gamma=\frac{3}{2}$.
Since, $f=3$ for monatomic gas and $f=5$ for diatomic gas.
Therefore, $\gamma_{1}=\left(1+\frac{2}{f}\right)=\left(1+\frac{2}{3}\right)=\frac{5}{3}$
$\Rightarrow \gamma_{2}=\left(1+\frac{2}{5}\right)=\frac{7}{5}$
where $f$ is number of degrees of freedom.
Now, substitute $n_{2}=n$ in Eq. (1), we get
$\frac{(2+n)}{\left(\frac{3}{2}-1\right)}=\frac{2}{\left(\frac{5}{3}-1\right)}+\frac{n}{\left(\frac{7}{5}-1\right)}$
$\frac{(2+n)}{1 / 2}=\frac{2}{2 / 3}+\frac{n}{2 / 5}$
$\Rightarrow 2(2+n) =3+\frac{5}{2} n$
$4+2 n=3+2.5 n$
$\Rightarrow 0.5 n=1$
$\Rightarrow n=2$