Solution:

The answer is (B). A number that in base $3$ ends with zero is a multiple of $3$; since no power of $4$ is a multiple of $3$, the numbers we are looking for, in base $3$, end with $1$. If we use exactly one digit $1$, the only possibility is then the number written as "1" in base $3$, that is $1$. Allowing another digit $1$, which corresponds to some $3^{k}$, means instead solving the equation $3^{k}+1=4^{a}$. Writing $4^{a}=2^{2a}$ we find $3^{k}=(2^{a}+1)(2^{a}-1)$, so both $2^{a}+1$ and $2^{a}-1$ are powers of $3$, and it is clear that the only two powers of $3$ at distance $2$ are $1,3$, hence $a=1$, and the only other power of $4$ with the required property is exactly $4$.