Answer:
I think its C, even with the typo.
Answer:
m=image distance÷object distance
Answer:

Explanation:
= Mg ion = 
= F ion = 
q = Charge of electron = 
r = Distance between ions = 
k = Coulomb constant = 
Electrical force is given by
![F=-\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=-\dfrac{8.99\times 10^9\times 2\times 1.6\times 10^{-19}\times -1\times 1.6\times 10^{-19}}{[(0.072+0.133)\times 10^{-9}]^2}\\\Rightarrow F=1.09527\times 10^{-8}\ N](https://tex.z-dn.net/?f=F%3D-%5Cdfrac%7Bkq_1q_2%7D%7Br%5E2%7D%5C%5C%5CRightarrow%20F%3D-%5Cdfrac%7B8.99%5Ctimes%2010%5E9%5Ctimes%202%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%20-1%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%7D%7B%5B%280.072%2B0.133%29%5Ctimes%2010%5E%7B-9%7D%5D%5E2%7D%5C%5C%5CRightarrow%20F%3D1.09527%5Ctimes%2010%5E%7B-8%7D%5C%20N)
The attractive force is 
(a) -48.0 cm, diverging
We can use the lens equation:

where
f is the focal length
p = 16.0 cm is the object distance
q = -12.0 cm is the image distance (with a negative sign because the image is on the same side as the object, so it is virtual)
Solving for f, we find the focal length of the lens:


The lens is diverging, since the focal length is negative.
(b) 6.38 mm, erect
We can use the magnification equation:

where
y' is the size of the image
y = 8.50 mm is the size of the object
Substituting p and q that we used in the previous part of the problem, we find y':

and the image is erect, since the sign is positive.
(c)
See attached picture.