Cardinality of the set of all Hamel bases. If we look at $\mathbb{R}$ as a vector space over $\mathbb{Q}$, then a Hamel basis $B$ has cardinality $\mathcal{c}$.
What about the cardinality of the set of Hamel bases? Is it simply $2^{\mathcal{c}}$?

This has nothing to do with the Continuum Hypothesis (see below).
Yes, the cardinality of the set of all Hamel bases for $\Bbb R$ over $\Bbb Q$ is $2^c$. It's certainly no larger than $2^c$, and for the other inequality you could use a cheap trick: If $B$ is a basis and $f:B\to\{1,-1\}$ then $\{f(b)b:b\in B\}$ is a Hamel basis.

Below Proof that $|B|=c$: Say $|B|=\kappa$. Let $\mathcal F$ be the set of finite subsets of $B$. Then $$|\mathcal F|\le\kappa+\kappa^2+\dots
=\kappa+\kappa+\dots=\kappa.$$ For each $F\in\mathcal F$ the number of functions from $F$ to $\Bbb Q$ is $\aleph_0^{|F|}=\aleph_0$. Hence $$c\le\aleph_0\kappa=\kappa.$$
