In joining B and B using nested loop method, with A in outer loop two factors are involved.
i>No. of blocks containing all rows in A should be fetched
ii> No. of Rows A times no of Blocks containing all Rows of B
(in worst case all rows of B are matched with all rows of A).
In above ques, $|R|<|S|$
i> will be less when number of rows in outer table is less since less no of rows will take less no. of blocks
ii> if we keep $R$ in outer loop, no. of rows in $R$ are less and no. of blocks in $S$ are more
If we keep $S$ in outer loop, no of rows in $S$ are more and no. of blocks in $R$ are less.
In ii> block accesses will be multiplication and will come same in both cases.
So, i> will determine no of block accesses
So, answer is A.