Step 1: Understand the test for magnetism.
A spin-only magnetic moment only appears when an ion carries at least one unpaired electron, since $\mu = \sqrt{n(n+2)}$ vanishes when $n = 0$.
Step 2: Decide which orbitals matter.
For first-row transition metal ions, only the partially filled $3d$ subshell decides the unpaired count, because the $4s$ electrons are removed first on ionisation.
Step 3: Examine $\text{Cu}^{2+}$.
Copper is $[\text{Ar}]3d^{10}4s^1$; removing two electrons leaves $3d^9$. A $d^9$ set has one lone electron, so $n = 1$ and it is magnetic.
Step 4: Examine $\text{Zn}^{2+}$.
Zinc is $[\text{Ar}]3d^{10}4s^2$; removing the two $4s$ electrons leaves $3d^{10}$. Every $d$ orbital is doubly filled, so $n = 0$ and there is no spin-only moment.
Step 5: Examine $\text{Ti}^{3+}$ and $\text{V}^{3+}$.
Titanium gives $3d^1$ ($n = 1$) and vanadium gives $3d^2$ ($n = 2$); both have unpaired electrons and are magnetic.
Step 6: Pick the odd one out.
Only $\text{Zn}^{2+}$ with its filled $d^{10}$ shell fails to show a spin-only moment, so the answer is option (2).
\[ \boxed{\text{Zn}^{2+}\ (d^{10},\ n = 0,\ \mu = 0)} \]