Record Size = 100 B
Block Size = 512 B
Hence, We can say, $\left \lfloor \frac{512}{100} \right \rfloor$ = $5$ records per block (Unspanned organisation assumed by default)
So, Number of block required for Database table = $\left \lceil \frac{50000}{5} \right \rceil = 10,000$ Blocks
Now, Let's find the Order $P$ of B+ tree.
Search Key size = 20 B
Pointer size = 12 B
And Order is defined as usual convention.
So, $(P-1)20 B + (P \times 12 B) \leq 512B$
Hence, $P = 16$
So, In any node of B+ tree, Minimum $\left \lceil \frac{P}{2} \right \rceil -1 = 7$ Keys should present.
Since, We are asked to make the Maximum index blocks in the B+ tree index, So, for that, We will try to fill each Index block as less as possible. Hence, We will try to put only $7$ keys per block.
Here, For B+ trees we can apply Bulk-loading concept. Since there are 10,000 blocks in the database relation.... For Sparse B+ tree index, We need to store 10,000 keys in the Index.
So, At the Bottom level (I am naming it 1st level) of Index, Putting 7 keys per node.. We need $\left \lfloor \frac{10000}{7} \right \rfloor = 1428$ Nodes.
Now, at next upper level (2nd level), We need $1428$ Node Pointers. And Since, We can put minimum 8 pointers per node, We would need $\left \lfloor \frac{1428}{8} \right \rfloor = 178$ Nodes
So, Now, at next upper level (3rd level), We need 178 Node pointers. Hence, $\left \lfloor \frac{178}{8} \right \rfloor = 22$ Nodes.
Going on same way, On next upper level (4th level), We need 22 Node pointers. Hence, $\left \lfloor \frac{22}{8} \right \rfloor = 2$ Nodes.
And finally, at the root level, We can have 2 Node pointers, Hence, 1 Block.
So, Maximum number of Index Blocks required = $1428 + 178 + 22 + 2 + 1 = 1631$ Blocks.
