Arens regularity of an ultrapower of an Arens regular Banach Algebra? Is an ultrapower of an Arens regular (non-superreflexive) Banach algebra, Arens regular? 
 A: I am not 100% sure what the question is asking.

Is the ultrapower of an Arens regular Banach algebra also Arens regular? 

As my comment said, if $A$ is a $C^*$-algebra, then "yes".
$\newcommand{\mc}{\mathcal}\newcommand{\ip}[2]{\langle #1,#2\rangle}$To show it's not always true, here's a counter-example.  Let $E$ be a reflexive Banach space, and let $A = \mc A(E)$ be the algebra of approximable (norm-closure of finite-rank) operators.  Then $A$ is Arens regular.  Let $\mc U$ be a non-principle ultrafilter on $\mathbb N$ (for example) and consider the ultrapower $(A)_{\mc U}$ which, in the obvious way, can be considered as a closed subalgebra of $\mc B((E)_{\mc U})$.  [The idea behind the following proof is that if $(E)_{\mc U}$ is not reflexive, then $\mc A( (E)_{\mc U} )$ is not Arens regular, and "morally" $\mc A( (E)_{\mc U} )$ is a subalgebra of $(A)_{\mc U}$, so $(A)_{\mc U}$ is not Arens regular.  This is not true, so we work a bit harder.]
I'll use Heinrich's wonderful paper Ultraproducts in Banach space theory  Sadly not open access (WHY?)  Namely, Corollary 7.5 which says:

Let $E_0\subseteq (E)_{\mc U}$ be a separable subspace.  For $\mu\in (E)_{\mc U}'$ there is $(\mu_i)\in (E')_{\mc U}$ with $\|\mu\| = \|(\mu_i)\|$ such that $\ip{\mu}{x} = \ip{(\mu_i)}{x} = \lim_{i\rightarrow\mc U} \ip{\mu_i}{x_i}$ for each $x=(x_i)\in E_0$.

We use this to cope with the fact that we'd like that $(E)'_{\mc U} = (E')_{\mc U}$ but this is only true if $E$ is super-reflexive.
Suppose that $E$ is reflexive but not super-reflexive, so that $(E)_{\mc U}$ is not reflexive.  Hence we can find sequences $(x^{(n)})$ in $(E)_{\mc U}$ and $(\mu^{(n)})$ in $(E)_{\mc U}'$ such that
$$ \lim_n \lim_m \ip{\mu^{(n)}}{x^{(m)}} \not= \lim_m \lim_n \ip{\mu^{(n)}}{x^{(m)}}. $$
That is, the iterated limits exist, but are not equal.  By the above result, we may suppose that each $\mu^{(n)} \in (E')_{\mc U}$.  For each $n$ let $x^{(n)} = (x^{(n)}_i) \in (E)_{\mc U}$ and $\mu^{(n)} = (\mu^{(n)}_i) \in (E')_{\mc U}$.  Fix $y\in E, \phi\in E'$ with $\ip{\phi}{y}=1$.  For each $n,i$ let
$$ a^{(n)}_i = \mu^{(n)}_i \otimes y, \qquad
b^{(n)}_i = \phi \otimes x^{(n)}_i. $$
Let $a^{(n)} = (a^{(n)}_i) \in (A)_{\mc U}$ and $b^{(n)} = (b^{(n)}_i) \in (A)_{\mc U}$. Define
$$ \psi : (A)_{\mc U}\rightarrow\mathbb C; \quad
\psi((c_i)) = \lim_{i\rightarrow\mc U} \ip{\phi}{c_i(y)}
\qquad ((c_i)\in (A)_{\mc U}). $$
Then
$$ \ip{\psi}{a^{(n)}b^{(m)}}
= \lim_{i\rightarrow\mc U} \ip{\phi}{a^{(n)}_i b^{(m)}_i(y)}
= \lim_{i\rightarrow\mc U} \ip{\mu^{(n)}_i}{x^{(m)}_i}
= \ip{\mu^{(n)}}{x^{(m)}}. $$
Hence the iterated limits of $\ip{\psi}{a^{(n)}b^{(m)}}$ are not equal, and so $(A)_{\mc U}$ is not Arens regular.
