For monoatomic gas;
$C_v = \large\frac{3}{2}RT$
$C_p =\large\frac{5}{2}RT$
For diatomic gas
$C_v = \large\frac{5}{2}RT$
$C_p = \large\frac{7}{2}RT$
Thus for mixture of 1 mole each
$C_v = \large\frac{\large\frac{3}{2}RT+\large\frac{5}{2}RT}{2}$
$C_p = \large\frac{\large\frac{5}{2}RT+\large\frac{7}{2}RT}{2}$
$\large\frac{C_p}{C_v} = \large\frac{3RT}{2RT}$
= 1.5
Hence answer is (b)