= Solution
{c}
A <k-Sperner family> meets every <maximal chain in a Boolean lattice> in at most $k$ members. The random-chain calculation in part (a) therefore gives the generalized <LYM inequality> $L(\mathcal A)\leq k$. Define the proportion of rank $i$ that is present by
$$
x_i=\frac{|\mathcal A\cap X^{(i)}|}{\binom ni},\qquad 0\leq x_i\leq1.
$$
Then $\sum_i x_i=L(\mathcal A)\leq k$ and $w(\mathcal A)=\sum_i u(i)x_i$. This reduces the <weighted theorem for k-Sperner families> to allocating at most $k$ units of mass among ranks, each of capacity one.
To handle ties explicitly, let $t>0$ be the $k$th largest value among the $u(i)$, let $H=\{i:u(i)>t\}$, and put $h=|H|<k$. Then
$$
\begin{aligned}
w(\mathcal A)&=t\sum_i x_i+\sum_{i\in H}(u(i)-t)x_i-\sum_{u(i)<t}(t-u(i))x_i\\
&\leq tk+\sum_{i\in H}(u(i)-t)\\
&=\sum_{i\in H}u(i)+(k-h)t.
\end{aligned}
$$
The last expression is exactly the sum of the $k$ largest rank weights, with multiplicity. Writing those values in descending order as $u_{(1)},\ldots,u_{(n+1)}$, the answer is
$$
\boxed{w(\mathcal A)\leq\sum_{j=1}^k u_{(j)}.}
$$
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