Kunen"s "Set Theory" on page 21, defines the Russell set as $R:\{x:x\notin x\}$.
In the context of $ZFC$, not set can be an element of itself.
Then would $R$ be a proper class since it would include all sets and not be a set by virtue of Burali-Forti?
Thanks
$R$ is indeed a proper class.
A simple, self-contained proof of this fact is the argument of Russell's paradox, which derives a contradiction from the hypothesis that $R$ is a set.
A virtue of this proof is that it remains valid in Z; the axiom of foundation plays no role in the argument.
Commonly, the fact $R$ is a proper class is used to prove the class of all sets is proper, since if there was a set of all sets, $R$ must also be a set by the axiom of subsets.