The electron is defined in the Standard Model as a elementary particle, pointlike, with no size or spatial extension.
Protons and neutrons that make up the nucleus are on the other hand composite particles as defined in the Standard Model, and they do have spatial extension. Of course, these are made up of quarks, antiquarks, and gluons, (contrary to popular belief not just three, those are the valence quarks only), but in reality a sea of quarks, antiquarks and gluons.
So based on this, it would not be correct to say that the nucleus, a composite particle, is smaller then the electron, a point particle, with no spatial extension.
However, what you are referring to, is the fact that, as per QM, the electron cloud is the physical manifestation of the mathematical description of something we call the probability distribution of the electron. This does have a spatial extension.
The diameter of the nucleus is in the range of 1.7566 fm (1.7566×10−15 m) for hydrogen (the diameter of a single proton) to about 11.7142 fm for uranium. These dimensions are much smaller than the diameter of the atom itself (nucleus + electron cloud), by a factor of about 26,634 (uranium atomic radius is about 156 pm (156×10−12 m)) to about 60,250 (hydrogen atomic radius is about 52.92 pm).[a]
And in our universe, it happens so, that the (radius of the) spatial extension of this electron cloud is bigger than the average radius of the nucleus (which does have a spatial extension too). So in this sense you could argue that your statement about the electron cloud and the nucleus is worth thinking about. So you could say this, the spatial extension of the nucleus (which is just the probability distribution of the sea of constituents) is smaller than the probability distribution of the electron (the cloud).