Knowing the rate of growth of various class of functions, we can solve it in an easy way
let $n^{\frac{1}{\sqrt {log_2n}}}=A$
$\sqrt{log_2n}=B$
and $n^{\frac{1}{100}}=C$
A and C are polynomial functions
B is poly-logarithmic function raised to power 1/2.
Polynomials always grow faster than poly-logarithms so, $B$ will be slowest of all.
Among A and C, as n will increase $\frac{1}{\sqrt{log_2n}}$ will decrease and hence A will grow more slowly as compared to C. But, power term in C is constant, so C will grow more fast with increasing n.
so A<C
required order B<A<C and this corresponds to option (D)