Lets identify the function types
- $f_1=10^{n}-$ Exponential Function
- $f_2=n^{\log n}-$ Super-polynomial but sub-exponential
- $f_3=n^{\sqrt n}-$ Super-polynomial but sub-exponential
So, clearly asymptotic growth of $f_1$ is the highest.
Both $f_2$ and $f_3$ are super polynomial (grows faster than any polynomial function) but sub-exponential (grows slower than any exponential function).
Now, $\log n$ grows slower than any power function for positive power (not necessarily a polynomial function). i.e., $\log n = o(n^x), x>0$. So, asymptotic growth of $\log n$ is lower than even $n^{0.000001}.$ $\sqrt n$ is a power function with power $ = 0.5.$ So, clearly $\log n = o(\sqrt n) \implies n^{\log n} = o \left(n^{\sqrt n}\right).$
So, asymptotic growth rate of $f_1,f_2$ and $f_3 $ are in the order $f_2<f_3<f_1.$
Option $D$ is correct.