For example, you have an Aircraft being subjected to a Load Factor of
2 and whose total mass is 5000kg and wings + fuel mass is 1500kg, and
are asked to calculate the Inertia Loading of Said wings.
Would you multiply the weight of the wings by 2?
Yes.
Note that, strictly speaking, kg are not a measure of weight. You would need to multiply the weight of the wings by two.
In more detail:
In the most general terms, "load factor" could be said to be equal total aerodynamic force divided by weight. The usual convention is to use "load factor" to mean lift force divided by weight. The difference between the two results could be very significant in a steep steady-state climb, or in a sideslip.
For the remainder of this answer, "load factor" will be used to mean lift force divided by weight.
As other answers have noted, just knowing the "load factor" is not enough to predict the stresses and strains at the wing root etc-- the load distribution matters as well.
It may help to note the following:
"Centrifugal force" is not a real force, in the sense that it is often applied to an analysis of the flight of a turning aircraft.
Gravitational force (weight) acts with an equal force per unit mass on every single particle of the aircraft, and thus exerts no stresses or strains on the aircraft structure.
However, it is not uncommon to hear the "inertial loading" acting on the aircraft as a whole being described as the combined vector sum of "centrifugal force" and gravitational force. Whether this is truly "accurate" or "inaccurate" may to some extent be a matter of opinion or reference frame; at any rate the net result is always exactly equal in magnitude (but opposite in direction) to the net aerodynamic force being generated by the aircraft. In the more narrow case where all aerodynamic forces other than lift add up to zero, the sum of "centrifugal force" and gravitational force will be equal in magnitude and opposite in direction to the lift vector. In other words, in the case where the net aerodynamic force is equal to the lift vector, the "inertial loading" acting on the aircraft as a whole is exactly equal in magnitude to the lift vector, which is exactly equal in magnitude to the "load factor" times the aircraft weight.
(Note that this conception of "inertial loading" is a little limited-- it considers effects related to turning but not effects related to rotation-- the difference will rarely be significant, except in unusual cases like a flat spin with a high rotation rate-- these nuances only apply to a discussion of the inertial load acting on a given element of the aircraft, not on the aircraft as a whole.)
It would seem that the original question is using the phrase "inertial loading of the wings" to describe the portion of the wing's lift force that is absorbed by accelerating the mass of each wing, and therefore does not contribute to the upward bending moment acting at the wing root. This would be equal to the weight of each wing, times the "load factor", with the "load factor" being defined as total lift force divided by total weight.
Since we don't know that the center of lift of each wing is the same as the center of mass of each wing, it's only a rough approximation to say that the lift "absorbed" by the "inertial loading" of each wing, calculated as we've stated above, makes no contribution to the bending moment at the wing roots.