See here for definition of Bing's space. There's also Dan Ma's blog or Counterexamples in Topology by Steen and Sebach.
Since the Michael's closed subspace $Y\subseteq X$ of Bing's space $X$ is metacompact but not paracompact, $Y$ is not countably compact. Therefore, $X$ is not countably compact. Since $X$ is normal, it's not pseudocompact.
Is $X$ realcompact? The map $f:X\to \{0, 1\}^{2^\mathbb{R}}$, $f(x) = x$, is a continuous bijection. If we were to show that every subspace of $\{0, 1\}^{2^\mathbb{R}}$ is realcompact, then this would imply that $X$ is realcompact.
Edit: We know that $Y$ is realcompact from other results (it's a $T_4$ metacompact space of non-measurable size).
If we could show that $X$ is subparacompact, or more generally $\theta$-refinable, then it would follow that $X$ is realcompact.
This would hold if we could show that $X$ has a $\sigma$-locally finite network, for then it'd be a $\sigma$-space, hence subparacompact (Bing's space H is subparacompact for this reason).
Edit 2: Note that the Bing's space H was specifically constructed so all the points in it are $G_\delta$. This means we probably can't show that Bing's space G has such network after all, since all non-isolated points are not $G_\delta$.
Edit 3: In the literature, $\theta$-refinable spaces are called submetacompact spaces. Here $X$ is a space which isn't collectionwise normal, so it's possible that it's submetacompact.
Edit 4: Another author shows that if in the construction of Bing's space G, the set we use instead of $\mathbb{R}$ is of size $>$ continuum, then space such obtained won't be submetacompact, using some results by Erdos. This doesn't mean above example isn't submetacompact, since we are interested in construction based on taking $\mathbb{R}$, a set of precisely size continuum.
Edit 5: While I'm not certain if the Bing's space constructed from $\mathbb{R}$ is $\theta$-refinable (or submetacompact), it's weakly $\theta$-refinable and shrinking. In particular it's countably paracompact.
Edit 6: I've missed this before, but $\{0, 1\}^{2^\mathbb{R}}$ is not hereditarily realcompact since $\omega_1$ embedds into it, and $\omega_1$ is not realcompact. It's not Borel-complete either since $\{0, 1\}^{\aleph_1}$ is not Borel-complete.
Since $X$ is normal and countably paracompact, if $X$ is Borel-complete then $X$ is realcompact. Since $X$ is not countably compact, $\beta X$ is not Borel-complete. Perhaps one could show $X$ is Borel-complete?