1. Write the electronic configuration of neutral Ho

Holmium has atomic number $67$.

So, the electronic configuration of $Ho$ is: $[Xe] \, 4f^{11} 6s^2$

2. Form the $Ho^{3+}$ ion

To form $Ho^{3+}$, remove $3$ electrons from the neutral atom.

In lanthanides, electrons are removed first from $6s$ and then from $4f$.

So:

• Remove $2$ electrons from $6s^2 \rightarrow 6s^0$
• Remove $1$ electron from $4f^{11} \rightarrow 4f^{10}$

Thus, $Ho^{3+} = [Xe] \, 4f^{10}$

3. Count electrons in the $4f$ orbital/subshell

The configuration $4f^{10}$ means there are $10$ electrons in the $4f$ subshell.

Therefore, the number of electrons present in $4f$ of $Ho^{3+}$ is: $\boxed{10}$