Options are ....
I) 20,30
ii) 30,40
iii)40,30
iv) 30,20
molar mass of B2A3 = 9/0.05 = 2B + 3A =180
molar mass of B2A = 10/0.1 = 2B+ A = 100
here A = 100 - 2B
subustitute the value of A in 1st eq., we get , 2B +3 (100-2B) = 180
2B +300 -6B = 180
here w get B =30
put the value of B in eq 2nd we get A = 40
hence, option (iii)
is correct
Study 40% syllabus and score up to 100% marks in JEE