Which of the following forms stable +4 oxidation state?

The stability of any oxidation state in lanthanoids depends on the electronic configuration obtained after the required number of electrons are removed. Closed-shell or exactly half-filled subshells impart extra stability.

Step 1 : Write ground-state configurations
La (57): $$[Xe]\,5d^{1}\,6s^{2}$$
Ce (58): $$[Xe]\,4f^{1}\,5d^{1}\,6s^{2}$$
Eu (63): $$[Xe]\,4f^{7}\,6s^{2}$$
Gd (64): $$[Xe]\,4f^{7}\,5d^{1}\,6s^{2}$$

Step 2 : Examine the species obtained on forming the +4 state
La⁴⁺ : four electrons must be removed but La has only three valence electrons. The fourth electron would have to come from the inner $$[Xe]$$ core, giving a highly unstable configuration. Hence +4 is not feasible for La.

Ce⁴⁺ : removal of 4 electrons (two from 6s, one from 5d, one from 4f) gives exactly the noble-gas core $$[Xe]$$ (i.e. $$4f^{0}\,5d^{0}\,6s^{0}$$). A complete shell is exceptionally stable, so Ce readily forms the +4 oxidation state (e.g. $$CeO_2$$).

Eu⁴⁺ : after removing four electrons $$[Xe]\,4f^{7}\,6s^{2} \rightarrow [Xe]\,4f^{6}$$. The half-filled $$4f^{7}$$ subshell of Eu²⁺/Eu³⁺ is lost, giving lower stability; therefore Eu prefers +2 (half-filled 4f⁷) and +3 states, not +4.

Gd⁴⁺ : $$[Xe]\,4f^{7}\,5d^{1}\,6s^{2} \rightarrow [Xe]\,4f^{6}$$ after losing four electrons. Again the half-filled $$4f^{7}$$ is destroyed, so +4 is disfavoured. Gd is most stable in the +3 state (retaining 4f⁷).

Conclusion
Only cerium achieves a noble-gas configuration in the +4 state, making that oxidation state stable. None of the other given lanthanoids obtain a specially stable arrangement on losing four electrons.

Option C which is: $$Ce (Z = 58)$$