When you will naturally join S and T both have E as common. As for S E is simple candidate key and will have unique values and in T E can take same values as it not simple. So,maximum we can get by joining is 80 tuples. Now, Joining with table R.as A is candidate key it will have unique value and in previous S join T there aslo A will have unique value, So, maximum tuple we will get will be 80. Now,Joining these three R,S,T with U common attribute will be D,G both are non-key so,maximum we get when both D,G are same for every table. So,80*10=800.
