6. Let the set $A=\{0,1,2, \cdots, 9\},\left\{B_{1}, B_{2}, \cdots, B_{k}\right.$ be a family of non-empty subsets of $A$, and when $i \neq j$, $B_{i} \cap B_{j}$ has at most two elements. Then the maximum value of $k$ is $\qquad$
6.175 .

It is not hard to see that the family of all subsets of $A$ containing at most three elements meets the conditions of the problem, where the number of subsets is $\mathrm{C}_{10}^{1}+\mathrm{C}_{10}^{e}+\mathrm{C}_{10}^{3}=175$.

Suppose there is another family of subsets $C$ that meets the conditions, and there is a subset $B \in C$, where subset $B$ has more than 3 elements. Let $a \in B$, since $B \cap (B \setminus \{a\})$ has more than 2 elements, hence $B \setminus \{a\} \notin C$. Then, when replacing $B$ in $C$ with $B \setminus \{a\}$, the new family of subsets still meets the conditions of the problem (since $B \setminus \{a\}$ has one fewer element than $B$). Repeating this process, we can eventually ensure that each subset in the resulting family has no more than 3 elements. Note that the number of subsets in the family does not increase after each replacement, so the number of subsets in $C$ is $\leqslant 175$.