The width of the poset \(\bftwo^t\) is at least \(C(t,\lfloor\frac{t}{2}\rfloor)\) since the set of all \(\lfloor\frac{t}{2}\rfloor\)-element subsets of \(\{1,2,\dots,t\}\) is an antichain. We now show that the width of \(\bftwo^t\) is at most \(C(t,\lfloor\frac{t}{2}\rfloor)\text{.}\)
Let \(w\) be the width of \(\bftwo^t\) and let \(\{S_1,S_2,\dots, S_w\}\) be an antichain of size \(w\) in this poset, i.e., each \(S_i\) is a subset of \(\{1,2,\dots,t\}\) and if \(1\le i\lt j\le w\text{,}\) then \(S_i\nsubseteq S_j\) and \(S_j\nsubseteq S_i\text{.}\)
For each \(i\text{,}\) consider the set \(\cgS_i\) of all maximal chains which pass through \(S_i\text{.}\) It is easy to see that if \(|S_i|=k_i\text{,}\) then \(|\cgS_i|=k_i!(t-k_i)!\text{.}\) This follows from the observation that to form such a maximum chain beginning with \(S_i\) as an intermediate point, you delete the elements of \(S_i\) one at a time to form the sets of the lower part of the chain. Also, to form the upper part of the chain, you add the elements not in \(S_i\) one at a time.
Note further that if \(1\le i \lt j\le w\text{,}\) then \(\cgS_i\cap \cgS_j =\emptyset\text{,}\) for if there was a maximum chain belonging to both \(\cgS_i\) and \(\cgS_j\text{,}\) then it would imply that one of \(S_i\) and \(S_j\) is a subset of the other.
Altogether, there are exactly \(t!\) maximum chains in \(\bftwo^t\text{.}\) This implies that
\begin{equation*}
\sum_{i=1}^{w} k_i!(t-k_i)!\le t!\text{.}
\end{equation*}
This implies that
\begin{equation*}
\sum_{i=1}^{w}\frac{k_i!(t-k_i)!}{t!}=
\sum_{i=1}^{w}\frac{1}{\binom{t}{k_i}}\le 1.
\end{equation*}
It follows that
\begin{equation*}
\sum_{i=1}^{i=w}\frac{1}{\binom{t}{\lceil\frac{t}{2}\rceil}}\le 1\text{.}
\end{equation*}
Thus
\begin{equation*}
w\le \binom{t}{\lceil\frac{t}{2}\rceil}\text{.}
\end{equation*}