Consider the below function:
int cuberoot(int n)
for(int i =1; i<=n;i++)
if (i*i*i<=n) crt=i;
The above function will return the cube root of value 'n' as defined in question.
Now if n = 64, the for loop will run 5 times(i=1,2,3,4,5) O(logn).
But if we take larger value say n= 8192(i.e.2^13) then the for loop will run 21 times which is (logn)^k , where k=1.1869
Therefore we can say that the complexity of computing cuberoot will O((logn)^k) but not O(loglogn).
Hence Answer is (C)