Put $n = 2^{16}$
i) $n log n \rightarrow 2^{16} log 2^{16} = 2^{16} . 16 = 2^{20}$
ii) $n log(log n) \rightarrow 2^{16} log (log (2^{16})) = 2^{16} . 4 = 2^{18}$
iii) $n log (n^n) \rightarrow n^2 logn = ({2^{16}})^2 log( 2^{16}) = 2^{32}.16 = 2^{36}$
iv) $n (logn)^2 \rightarrow 2^{16} (log (2^{16}))^2 = 2^{16}. 256 = 2^{16}. 2^8 = 2^{24}$
Therefore the correct increasing order is $ii < i < iv < iii.$
So, the answer is None of these.