DB size 512
record size 16
no of record possible in disk block 512/16=32 records/block
no of Disk block present in DB file 8192/32=256
so primary index is used which is called as sparse index .in this index we make entry of each disk block but not of every record.!
so at first level we have =256/(index file size )
now wt is index file : with the help of block pointer and search we locate reocrds so index file size =512/(6+10)
=512/16 =32 (records per index )
so at first level we have 256/32=8 index blocks
and hence multilevel index is used so in last level we have only 1 index block
so (d ) ans.