Let the number of electrons present in ion `A^(3+)` be `x`.
`:.` Number of neutrons in it `= x + 30.4%` of `x = 1.304 x`
Since the ion is tripositive,
`rArr` Number of electrons in neutral atom = x + 3
`:.` Number of protons in neutral atom = x + 3
Given,
Mass number of the ion = 56
`:. (x + 3) + (1.304x) = 56`
`2.304x = 53`
`x = (53)/(2.304)`
`x = 23`
`:.` Number of protons = x + 3 = 23 + 3 = 26
`:.` The symbol of the ion `._(26)^(56)Fe^(3+)`.