quantum number - question about the ms-value
As porphyrin clearly explained, the first three quantum numbers usually are enough to determine the orbital of a given electron.

You probably know this but I'd like to let it here for everyone else.

Each different value of the principal quantum number $n$ yields a different wave function and a different state of the electron. In atoms, $n$ defines the order of all states according to energy. 

The second and third quantum numbers $l$ and $m_{l}$, arise from the quantization of rotation where $l$ gives the shape of an orbital and its angular distribution and $m_{l}$ determines the number of orbitals and their orientations in a subshell.

The spin quantum number of an electron is denoted $s$ and it has the value of $s=1/2$. The spin magnetic quantum number is denoted $m_{s}$ gives the projection of the spin angular momentum on the z-axis. Therefore, one electron can have only one orientation: either $m_{s}=+1/2$ which we commonly denote up-spin or $\alpha$ and $m_{s}=-1/2$ denoted as down-spin or $\beta$.

What I see in the comments are partial answers because to associate an electron (not an orbital) to these numbers you need to use $m_{s}$. Then, you have a spin-down or $\beta$ electron in a $3d$ orbital.

Answering your question in the comments. To have a $3d^{2}$ you'd need two pairs of four quantum numbers: $(3,2,2,+1/2)$ and $(3,2,2,-1/2)$. Here, two electrons occupy the same region in space but they have different orientations with respect to their spin angular momenta.