I think its $O(n^2)$.
The inner most for loops:
for(int j=0;j<n;j+=2)
Takes $O(n)$
for(int j=1;j<n;j*=2)
Takes $O(logn)$
Since they are in sequence : $O(n) + O(logn) = O(n)$
Now evaluate outside in:
b 
1 
2 
4 
8 
.... 
.... 
.... 
.... 
$2^k = n$ 
i ( 0 to b) 
1 time 
2 times 
4 times 
8 times 




$2^k$ times 
innermost 
$O(n)$ 
$O(n)$ 
$O(n)$ 
$O(n)$ 




$O(n)$ 
Total 
n times 
2n times 
4n times 
8n times 




$2^k * n$ times 
Sum up the total number of times:
$ = n + 2n + 4n+ 8n+......+2^k *n$
$ = n (1 + 2 + 4 + 8 + ...... + 2^k)$
Also $2 ^k = n \implies k = log n$
$ = n (1 + 2 + 4 + 8 + ...... log n\:times)$
Apply the GP summation $ a*\frac{(r^p  1)}{r 1}$, here $a = 1, r = 2, p = log n$
$ = n*(1 *\frac{2^{logn}1}{21})$
$ = n*(1 *\frac{n1}{1})$
Which is clearly $O(n^2)$