Given maximum number of keys in any node is 5 ...
So Order of non-leaf node(maximum number of children) is 5+1 = 6
Order of leaf node(maximum number of keys) = 5 which is given in question...
Minimum number of keys possible in any non-leaf node = [ ceil (order_of_nonleaf_node) / 2 ] - 1 = 2
Minimum number of keys possible in any leaf node = [ ceil (order_of_leaf_node) / 2 ] = 3
So minimum number of nodes possible in the B+ tree = min (2,3) = 2
(we have ignored the special case of root as per question which can even have 1 key, EX :when our data file has only one record)