f(n) = 3n^√n
g(n) = 2^√nlog2n
here assume n=2^128
f(n)=log(3n^√n ) here 3 is just a constant to ignore it for a while so it will be n^√n now after taking log it will be √n log n now put the value of n so 2^64log2^128= 2^64*2^7= 2^71.
now g(n) = 2^√nlog2n here (log2n 2 is base of log)
after taking log both side √nlog2nlog2 after putting the value of n 2^64*log2^128*log2
which is again 2^71
here we are watching that both f(n) and g(n) are same and big O notation is valid for <= means f(n)=Og(n)
yes it can also g(n)=Of(n).
so option D is correct.