search field is V = 4 bytes long, the block size is B = 4096 bytes, and a block pointer is P = 8 bytes. As internal node of the B+ tree can have up to p tree pointers and p-1 search field values, these must fit into a single block. Hence, we have,
(p * 8) + (( p – 1 ) * 4 <= 4096
or (12 * p) <= 5000
p = 416.66 , So max keys are possible is 415 ( we cant consider the fraction )
But is this problem they done it this way
As internal node of the B+ tree can have up to p+1 tree pointers and p search field values, these must fit into a single block. Hence, we have,
(p+1) * 8 + p * 4 <= 4096
or (12 * p) <= 4088
p = 340.66 , So max keys are possible is 340 ( we cant consider the fraction )