For example, in the silicon atom, do the electrons located in $3s$ and $3p$, with $n=3$, have the same distance to the nucleus? I think in the diagram below, the arrangement of the electrons relative to the nucleus is based on energy, not the distance of the electron from the nucleus and electrons in specified orbitals (3s & 3p with $n=3$) are
the same distance from nucleus. Is this statement correct?
This is a quantum mechanical (QM) system, any attempt to think about it in purely classical ways is destined to fail to some extent.
In QM, the orbits of the electrons dot not have a fixed radius but have a certain probability amplitude to be found at a certain distance from the nucleus.
So a better question would be, what is the average radius of an electron in an atom.
This question is still hard for mosts atoms. As we do not have full analytical solutions. If we accept to approximate a silicon atom to a hydrogen-like system with 14 non-interacting electrons, we can provide some answers.
For hydrogen-like systems, the mean radius for a given eigenstate is given by
$$\langle r \rangle=\langle n ,\ell ,m_z|r|n,\ell,m_z\rangle=\frac{[3n^2-\ell(\ell+1)]a_0}{2Z}$$
where $a_0$ is Bohr radius and $Z$ is the atomic number. Here $n$ is the principal quantum number and $\ell$ corresponds to the azimuthal quantum number, for p-states it is $\ell=1$ for s-states it is $\ell=0$.
So the answer is no, 3p electrons are usually closer to the nucleus than 3s electrons if we look at $\langle r \rangle$. Note that the answer it is probabilistic and can be different if you instead ask for other mean quantities like $\langle 1/r \rangle$ or $\langle r^2 \rangle$.
