Polynomial Functions: $n^{1/4}, n^{1000}$
Exponential Functions: $e^n, 3^n, 1.001^n$
Factorial: $n!$
So, it is straightforward that
$n^{1/4} < n^{1000} < 1.001^n < e^n < 3^n < n!$
We only, need to place $n^{\sqrt n}.$
Lets compare $n^{\sqrt n}$ with ${1.001}^n$
By logarithm rule, $n^{\sqrt n} = 1.001^{\log_{1.001}n}\sqrt n = 1.001 ^{\sqrt n \log_{1.001} n} $
Now, comparing the exponent parts, we get
$\sqrt n \log_{1.001} n$ and $n$
$\implies \log_{1.001}$ and $\sqrt n.$
Now, $\log n < \sqrt n$ (Proof see below).
So, we get $n^{\sqrt n} < 1.001^n.$
So, the correct order is
$n^{1/4}< n^{1000}<n^{\sqrt n}<1.001^n <e^n< 3^n< n!$
Option D.
$\log n = 2 \log n^{0.5} = 2 \log \sqrt n < \sqrt n $