10 mole of ${SO}_2$ and 15 mole of $O_2$ were passed over a catalyst to produce 8 mole of ${SO}_3$. The ratio of ${SO}_2$ and ${SO}_3$ moles in mixture is

Balanced chemical reaction

$2{SO}_2+O_2\to 2{SO}_3$

Initial concentration of $O_2$ = 10 mole and $O_2$ = 15 mole

At equilibrium, let x mole of $O_2$ is consumed 

Therefore, 2x of ${SO}_3$ will form

Mole of ${SO}_3$ formed is given = 8 mole

$2x=8\text{ or }x=4$

The ratio of ${SO}_2$ and ${SO}_3$ moles in mixture is (10-8)/8=1/4. 

Therefore, option B is correct.