There can be no set of all sets that are members of themselves that is itself a member of itself (just as there can be no set of all sets that are not members of themselves that is itself not a member of itself).
Proof of the above:
When I write x (x, y, z), I mean to say set x contains elements x, y, and z. With this in mind, consider the following:
x (x, y, z): Here, if I say x is a set that is not a member of itself, we get a contradiction. (because x is in x)
x (x, y, z): Here, if I say x is a set that is a member of itself, we get no contradiction (because x is in x)
x (x, y, z): Here, if I say x is a set of three sets that are members of themselves, we get a contradiction (it amounts to x being a member of itself twice because x is in x. Either y and z are not members of themselves (making x the only member of itself), or x is a member of itself twice whilst y and z are members of themselves once (which is contradictory). It cannot be the case that x is a set of three sets that are members of themselves).
x (p, q, y): Here, x may be a set of three sets that are members of themselves precisely because it does not include itself. This shows that a set of all sets that are members of themselves, that is itself a member of itself, is impossible (the notion of any set containing more than one set as a member of itself is contradictory).
Russell’s paradox arises from the belief that you cannot have a set of all sets that are not members of themselves that is itself not a member of itself, whilst you can have a set of all sets that are members of themselves that is itself a member of itself. The above shows that this belief is contradictory (semantically inconsistent). Thus, paradox resolved.