See here search key value of a B-tree is k=8B
disk block size is B=512B,
record(data) pointer is P=6B
block pointer is i=5B so as per our knowledge we know
n*i+(n-1)*(k+Pr)<= disk block size
so
n*5+(n-1)*(8+6)<=512
=> n <=27.68
=> n will give 27
1ans means no. of node will be 27
and so no. of key will be 26 // here sometimes we forget to do that (no. of key = node no. - 1)
now since each node of B tree is 69 percent full
2ans therefore the average number of keys per node of the above tree
= 0.69 ˟ 26 =17.94 and approx we will take it 17
hope it is clear