This is related to yesterday's question Constructing a vector space of dimension $\beth_\omega$; it's the next exercise (I.13.35 (a)) in Kunen's Set Theory.
Let $B_0 = \ell^1$ and let $B_{n+1} = B_n^{**}$ be the continuous second dual of $B_n$, so that we can consider $B_n$ as a subset of $B_{n+1}$ in the usual way. Let $B$ be the completion of $\bigcup_n B_n$. Show that $|B| = \beth_{\omega+1}$.
I have managed to solve part (b) of the exercise, which says that any Banach space $X$ with $|X| \ge \beth_\omega$ actually has $|X| \ge \beth_{\omega+1}$. So it will be enough to show $|B| \ge \beth_\omega$ (though given the ordering of the parts of the exercise, this may not be what Kunen had in mind).
Presumably we should try to show that $|B_{n+1}| \ge 2^{|B_n|}$ or something similar. But when working with continuous duals, I don't see how to do that. Additionally, each step of the induction is somehow going to have to use the fact that we started with $B_0 = \ell^1$, since if $B_0$ had been a reflexive space, this would never work. We also have to rule out the possibility that one of the later $B_n$ turns out to be reflexive.
Any suggestions are welcome.
Let us try to estimate $|B|$ from below. Every space $B_{n-1}^*$ ($n\geqslant 1)$ is of the form $C(K_n)$ for some compact Hausdorff space. For example, $K_1 = \beta\mathbb{N}$ and $|K_1| = \beth_1$. In particular, by the Riesz–Markov–Kakutani representation theorem each space $B_n$ is isometric to the space $M(K_n)$ of Radon measures on $K_n$. Moreover, we can embed injectively $K_{n}$ ($n\geqslant 1$) into $B_n$ via $$x\mapsto \delta_x\;(x\in K_n).$$ The subspace $D_n:=\overline{\mbox{span}}\{\delta_x \colon x\in K_{n}\}\subset B_n$ is isometric to $\ell_1(K_{n})$.
We shall prove inductively that $|B_{n}|\geqslant \beth_{n}$ for each $n\geqslant 2$.
Suppose that $|D_n|\geqslant \beth_{n}$. It is clear that $|B_n|\geqslant |D_n|=|\ell_1(K_n)|=|K_n|$. In particular, $|K_n|\geqslant \beth_n$.
We claim that $|D_{n+1}|\geqslant \beth_{n+1}$. Indeed, $$|B_n^{**}|\geqslant |D_n^{**}|=|C(\beta |K_{n}|)^{*}|\geqslant |\beta |K_{n}||\geqslant 2^{\beth_{n}}=\beth_{n+1},$$ where the last inequality follows from Pospíšil's theorem.
We have thus proved that $|B|\geqslant \beth_\omega$. However, if $\kappa$ is the cardinality of a Banach space, then $\kappa^{\aleph_0}=\kappa$. Consequently, $|B|\geqslant (\beth_\omega)^{\aleph_0}=\beth_{\omega+1}$ (see this thread for the proof).
The opposite inequality is quite easy.