No. of blocks in disk = (30000x150)/512 = 8789.
In primary indexing , one entry for each block is there , so number of blocks in index = (8789 x 16)/512 = 274.65 = 275.
Now , we can apply binary search for primary indexing .
ceil(log(275)) = 9
In worst case , it'll take 9 block transfers to locate the exact block and 1 extra transfer for getting that block.
Therefore total number of transfers = 10.