Step 1: Count from lanthanum:
$\text{La}^{3+}$ is $4f^0$ (Z = 57). Each next element adds one 4f electron in the +3 state.
Step 2: Place the options:
Ce (58): $f^1$. Dy (66): $f^9$. Tm (69): $f^{12}$. Lu (71): $f^{14}$.
Step 3: Conclusion:
The 4f subshell holds at most 14 electrons, and the last element Lu completes it. So Lu is the one.
Step 4: Link to the lanthanoid contraction:
Across the series each added electron goes into the inner 4f subshell. That is why the +3 ions get steadily smaller, and why Lu3+ with a full 4f subshell is the smallest and most stable. The cerium, dysprosium and thulium ions have partly filled 4f subshells, so they cannot show f14.
Final Answer:
Option (B), lutetium.
\[ \boxed{\text{Lu (B)}} \]