For $2{\mathrm{SO}}_2+{\mathrm{O}}_2\to 2{\mathrm{SO}}_3$ rate of disappearance of $S{O}_2$ is $4\times {10}^{-3}{\mathrm{Ms}}^{-1}$ at t = 10 sec.  Then amount of $S{O}_3$formed and amount of $O_{2 }$consumed at t = 10 sec respectively are

As per reaction $2{\mathrm{SO}}_2+{\mathrm{O}}_2\to 2{\mathrm{SO}}_3$ 
Rate of disappeaerance of $S{O}_2 =2$ times the  rate of disappeaerance of $O_2 =$Rate of appeaerance of $S{O}_3$
Amount of $S{O}_{3 }$formed at 10th sec $= 4\times {10}^{-3}\times 80=0.32 g$
Amount of $O_{2 }$consumed at 10th sec $=4\times {10}^{-3}\times 32/2=0.064 \mathrm{g}$