Lets count the number of iterations which gives us the time complexity :
When i = 1 , m = floor(n/1) = n
when i = 2 , m = floor(n/2) which can be written as n/2 for complexity calculation convenience
when i = 3 , m = floor(n/3) which we write n/3
..... and so on till
i = n , m = floor(n/n) = 1
So no of times the program runs..For that we consider no of times the inner loop is run which is defined by the value of m for each iteration.
Hence time complexity = n[ 1 + 1/2 + 1/3 .........+ 1/n]
= θ(nlogn) [ Since 1 + 1/2 +.........1/n = logn ]
Hence A) option is true