The initial force between the two charges is given by:

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:
1. F
In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.
So, we have:

So, the new force is:

So the force has not changed.
2. F/4
In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.
So, we have:

So, the new force is:

So the force has decreased by a factor 4.
3. 6F
In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.
So, we have:

So, the new force is:

So the force has increased by a factor 6.
Explanation:
By Hooke's Law,
F=kx
The only force acting here is weight, and x is the extension of the string (you need to convert this to mm) so
mg=kx
(0.15)(9.81)=k ((420-300)x10^-3)
Then just solve this equation.
B) a new element is formed
Answer:
I would love to help, Could you put the question in English?
Explanation: