Here we've to eliminate options one by one.
check 1st letter of every option,there we've a confusion about A & D, so now let's compare between A & D
$n^{1/3}$ & $nlog^{9}n$ $\Rightarrow$ $\frac{logn}{3}$ & logn + 9loglogn
So clearly D is bigger than A, i.e option B is out.
Now let's take next two letter in the remaining options d,c. here we've a confusion b/w d & c,
$nlog^{9}n$ & $n^{7/4}$ $\Rightarrow$ $log^{9}n$ & $n^{3/4}$
take n = $10^{100}$ then $n^{3/4}$ = $10^{75}$ & $log^{9}n$ = $10^{18}$ i.e D<C
Now we can eliminate option C & D also and remaining option A is the answer.
If we Compare B & E , $e^{n}$ & $1.0000001^{n}$ (take log base e on both side)
It becomes n & n.$\ln(1.0000001)$ which clearly shows B>E
Finally a, d, c, e, b which is option A