The internal energy of $n$ moles of an ideal gas at temperature T is given by

      $\mathit{\text{U}}=\frac{f}{2}\mathit{\text{nRT}}$     ($f=$ degrees of freedom)

      $U_1=U_2$

$f{}_1{n}_1{T}_1=f{}_2{n}_2{T}_2$

$\therefore \frac{n_1}{n_2}=\frac{f{}_2{T}_2}{f{}_1{T}_1}=\frac{3\times 2}{5\times 1}=\frac{6}{5}$$=1.2$

Here, $f_2$ = degrees of freedom of He = 3

and, $f_1$  = degrees of freedom of H₂ = 5