The hypothesis says exactly that $\mathcal A$ is a <k-Sperner family> with $k=2$: it contains no three members forming a strict inclusion chain. Every <maximal chain in a Boolean lattice> consequently contains at most two members of $\mathcal A$, so the same random-chain argument gives $\lambda(\mathcal A)\leq2$. For a fixed amount of <Lubell mass>, cardinality is maximized by using the levels with the two largest <binomial coefficients>. The <Erdős theorem on k-Sperner families> therefore gives
$$
\boxed{
|\mathcal A|\leq
\begin{cases}
\binom{2m}{m}+\binom{2m}{m-1},&n=2m,\\[2mm]
2\binom{2m+1}{m},&n=2m+1.
\end{cases}}
$$
The bound is attained by taking the union of two levels having those sizes.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 109 / 1 / i / Solution

{c}

Among $k$-Sperner subfamilies of $\mathcal P([n])$, the maximum cardinality is the sum of the $k$ largest <binomial coefficients> $\binom nj$. The union of the corresponding levels attains the bound. For the upper bound, the <Lubell mass> is at most $k$, and assigning its available mass to the levels with largest binomial coefficients maximizes cardinality.

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 Incoming links: Erdős theorem on k-Sperner families