We need to find the maximum no. of nodes to be accessed so consider the minimum fill factor.
Here,order of b+tree= #ptrs per node=p=100
Mini. ptrs per node=$\left \lceil \frac{p}{2} \right \rceil$=50
#Nodes in last level=106/50 = 2 * 104
#Nodes in Second last level= 2*104/50 =400
#Nodes in Third last level= 400/50 =8
#Nodes in Forth last level= 8/50 =1
The maximum no. of nodes to be accessed = #B+ tree levels = 4 Ans(b)