20.1.1 $\star \star A=\{0,1,2, \cdots, 9\}, B_{1}, B_{2}, \cdots, B_{j}$ are a family of non-empty subsets of $A$, when $i \neq$ $j$, $\left|B_{i} \cap B_{j}\right| \leqslant 2$, find the maximum value of $k$.
Parse the number of non-empty subsets containing at most three elements:
$$
\mathrm{C}_{10}^{1}+\mathrm{C}_{10}^{2}+\mathrm{C}_{10}^{3}=175 \text {  subsets. }
$$

Let $C$ be the family of subsets that satisfy the condition. If $|B| \geq 4$, then $B$ and $B-\{b\}(b \in B)$ cannot both be contained in $C$. By replacing $B$ with $B-\{b\}$, the number of elements in $C$ remains unchanged. Similarly, for all sets $B$ in $C$ with at least 4 elements, perform the corresponding replacements. After these replacements, each set in the family $C$ will be a non-empty set with at most 3 elements. Therefore, the maximum value of $k$ is 175.