La stands for Lanthanum, which is a member of the lanthanide series of elements in the periodic table. It has an atomic number of 57 and an electronic configuration of [Xe]5d1 6s2.

Achieving Noble Gas Configuration
To achieve a noble gas configuration, an element must have its outermost shell filled with electrons. The noble gases are stable because they have a filled outermost shell. Lanthanum can achieve this configuration by losing three electrons from its outermost shell.

Oxidation States
The oxidation state of an element is the charge that it carries when it forms a compound or ion. Lanthanum can form compounds in various oxidation states, including +2, +3, and +4.

In order to determine in which oxidation state La achieves noble gas configuration, we need to look at its electron configuration and determine how many electrons it needs to lose to achieve a filled outermost shell.

La3+ (Oxidation State of +3)
When Lanthanum loses three electrons, it forms a +3 ion (La3+). In this oxidation state, La has a noble gas configuration of [Xe]. Therefore, option B is the correct answer.

Other Oxidation States
Lanthanum can also form compounds in other oxidation states, but they do not achieve a noble gas configuration. For example, in the +2 oxidation state, La has an electron configuration of [Xe]5d1, which is not a filled outermost shell.

Conclusion
In conclusion, Lanthanum achieves noble gas configuration in the +3 oxidation state by losing three electrons and forming a La3+ ion.