Hi,
if n has p factors, 2n has 2p factors..
if n = (a^i).(b^j).(c^k)... where a,b,c are all distinct primes.
So, p = (i+1)(j+1)..
Now 2n = 2^1.(a^i).(b^j).(c^k)...
Number of factors of 2n will be (1+1).(i+1)(j+1)... For this to happen, a,b,c... should be distinct from 2. So, n is product of powers of primes other than 2 i.e. product of odd numbers. So, n is odd number.
Sufficient

Hence, B

You can even follow this link:
https://www.beatthegmat.com/n-odd-t84998.html