$\text{Relation}\ R \ \text{require} \ \frac {1200}{100} = 12 \ \text {blocks}$
$\text{Relation}\ S \ \text{require} \ \frac {1000}{100} = 10 \ \text {blocks}$
$\text{Relation}\ R \ \text{has #tuples} = \ \frac {1200}{20} = 60 \ \text {tuples}$
$\text{Relation}\ S \ \text{has #tuples} = \ \frac {1000}{20} = 50 \ \text {tuples}$
Since $S$ is smaller thus it will be chosen as inner relation. In the worst case, one block of each relation can fit in memory, so the number of block transfers requires for nested join is $(50 × 12) + 10 = 610 $ and for block nested join is $(12 × 10) + 10 = 130$
$\text {Therefore,}$ if we use a nested join instead of a block nested join we will require $610− 130 = 480$ extra block transfers.
$\text {So, ans is 480}$
Reference
Question 4
https://gateoverflow.in/147193/nptel