To find the number of d-electrons in \( \text{Cr}^{3+} \), we need to first determine the electron configuration of the neutral chromium atom (\( Z = 24 \)).
Chromium has an atomic number of 24, so its electron configuration in the ground state is:
\[
\text{Cr}: [Ar] \, 3d^5 \, 4s^1
\]
When chromium loses 3 electrons to form the \( \text{Cr}^{3+} \) ion, the electrons are removed first from the 4s orbital and then from the 3d orbital. Thus, the electron configuration of \( \text{Cr}^{3+} \) is:
\[
\text{Cr}^{3+}: [Ar] \, 3d^3
\]
Therefore, the number of d-electrons in \( \text{Cr}^{3+} \) is 3.
Thus, the correct answer is \( \boxed{3} \).