Use the linear algebra method: For each $b \in B$, define $f_{b}: Z_{3}^{n}\rightarrow \mathbb{Z}_{3}$ as follows: \[f_{b}(x_{1},\ldots,x_{n})=\prod_{i=1}^{n}(x_{i}-b_{i}-1).\] From the hypothesis we have that for all $b \neq b' \in B$, $f_{b}(b')=0$. Further $f_{b}(b)\neq 0$. By a simple linear algebra argument $\{ f_{b}$, $b \in B\}$ are all independent in the vector space of functions from $\mathbb{Z}_{3}^{n}$ to $\mathbb{Z}_{3}$. Let $\mathcal{U}$ be the linear span of $\{ f_{b}, b \in B\}$, so that $dim(\mathcal{U})=|B|$. Now, since each $f_{b}$ is a linear combination of monomials of the form $\prod_{i \in S}x_{i}$, $S \subset [n]$ and there are $2^{n}$ such monomials, $dim(\mathcal{U}) \leq 2^{n}$. Therefore $|B|\leq 2^{n}$.

hhaller wrote: $dim(\mathcal{U})=|B|$. What is this ? I don't understand