Correct Answer - `K^(o+) gt Mg^(2+) gt Al^(3+) gt Li^(o+)`
All of then are not `(("Charge")/("No. of elecrons"))` ratio, i.e. `((Z)/(e))`, the larger is the ionic radii.
For `Li^(o+), (Z)/(e^(-))` ratio `= (3)/(2) = 1.5`
`K^(o+), (Z)/(e^(-))` ratio `= (19)/(18) = 1.055`
`Mg^(2+), (Z)/(e^(-))` ratio `= (12)/(10) = 1.2`
`Al^(3+), (Z)/(e^(-))` ratio `= (13)/(10) = 1.3`
So, the order of decreasing ionic radii is as :
`K^(o+) gt Mg^(2+) gt Al^(3+) gt Li^(o+)`