Solution:

The answer is (C). Let $n$ be a positive integer. Let $a_{k}, a_{k-1}, \ldots, a_{0}$ be the $k+1$ digits, from left to right, of the representation of $n$ in base 2. Each of the $a_{i}$ will take a value between 0 and 1, and we may assume without loss of generality that $a_{k}=1$. Therefore $n=2^{k} a_{k}+2^{k-1} a_{k-1}+\ldots+a_{0}$. In particular the estimate $n<2^{k+1}$ holds. Suppose now that $n$ satisfies the hypotheses of the problem, and hence that $a_{k}, a_{k-1}, \ldots, a_{0}$ are also the digits of the representation in base 3 of $2 n$. Then we have $2 n=3^{k} a_{k}+3^{k-1} a_{k-1}+\ldots+a_{0}$ and the estimate $2 n \geq 3^{k}$ holds. Now if $k \geq 4$ we have $3^{k}>2^{k+2}$ and hence, putting together the estimates obtained above, we get $2 n \geq 3^{k}>2^{k+2}>2 n$, which is not possible. Hence $k \leq 3$. Now by hypothesis
$$27 a_{3}+9 a_{2}+3 a_{1}+a_{0}=2\left(8 a_{3}+4 a_{2}+2 a_{1}+a_{0}\right)$$
that is
$$11 a_{3}+a_{2}-a_{1}-a_{0}=0$$
from which one immediately deduces that $a_{3}=0$ (otherwise the expression on the left is greater than 8; the $a_{i}$ take the value 0 or 1). Now $a_{2}-a_{1}-a_{0}=0$ has exactly two solutions (with at least one digit different from 0), namely $(a_{2}, a_{1}, a_{0})=(1,0,1)$ and $(a_{2}, a_{1}, a_{0})=(1,1,0)$ which correspond respectively to the two numbers $n=6$ and $n=5$.