Relation R size = 600 B
Relation S size = 500 B
Size of tuple in both relation = 20 B
number of tuples in relation R ($n_{r}$) = 600/20 = 30
number of tuples in relation S ($n_{s}$) = 500/20 = 25
Block Size = 200 B
So we find the number of blocks in relation R ($b_{r}$) and S($b_{s}$)
$b_{r}$= 600/200 = 3 and $b_{s}$ = 500/200 = 2.5 = 3 approx
According to nested loop join concept In worst case it will involve $n_{r}$*bs+ br block transfers plus nr+br seeks.
Block transfers in nested loop join = 25 * 3 + 3 = 78 (In nested loop join, S will be in outer as it has min tuples so we take
$n_{s}$ in place of $n_{r}$ .For each record/tuple in S we'll scan a block of R)
According to Block nested loop join concept In worst case it will involve $b_{r}$ * $b_{s}$ + $b_{r}$ block transfers
plus 2 * br seeks
Block transfers in Block nested loop join = 3 * 3 + 3 = 12
Extra block transfers be required, in the worst case, if nested loop join is used instead of block nested loop join technique to compute R⋈S = 66
Hence option 3 is correct.