How many d-electrons are in the outer shell of metals like Cadmium (this is counting the 4d electrons)?
Without doing some sort of at least half-serious calculations, you cannot really answer the question. For all we know, a cadmium(II) ion could be $[\ce{Kr}]\,\mathrm{4d^{10}}$, $[\ce{Kr}]\,\mathrm{4d^9\,5s^1}$ or $[\ce{Kr}]\,\mathrm{4d^8\,5s^2}$.

However, given the analogies to other transition metals and the general structure of practically all transition metal ions, the first configuration, $[\ce{Kr}]\,\mathrm{4d^{10}}$ is most likely. I can’t recall a case where a transition metal ion had any s electrons in the valence shell before the d electrons were completely populated; i.e. even if you have two s electrons in the neutral atom and ionise only once you would end up with only d electrons in the ion.