Design a decay series for a nuclide?

Well, to find the atomic number, we add `A`, `B`, and `C` (whatever those are):

`20 + 14 + 20 = 54`

The sum of these digits is `9` (i.e. not two-digit, and is between `1` and `9`), so there is no need to repeat this. This number is added to `90` to give `Z = 99`, for `"Es"` (einsteinium).

To find the mass number, we add `D`, `E`, and `F` (whatever those are):

`22 + 02 + 15 = 39`

Since the result is over `30`, we subtract `30` until the result is `< 30`. We obtain `9`, then. Since the value is odd and not divisible by `4`, we add `1` to get `10`. Lastly, we add this value to `230` to get the mass number, `A_m = 240` (to distinguish it from `A = 20`...).

(If you don't follow, I am literally following the directions on the image.)

So, we currently have

`bb(""_(Z)^(A_m) X = ""_(99)^(240) "Es")`,

einsteinium-240. I'm guessing that's where you wanted me to stop.

Identify the starting nuclide

`"Atomic number = A + B + C" = 20 + 14 + 20 = 54 rArr 5 + 4 = 9`
`rArr 90 + 9 = 99`

`"Mass number = D + E + F" = 22 + 2 + 15 = 39 rArr 39 - 30 = 9`
`rArr 9 + 1 = 10 rArr 230 + 10 = 240`

So, we must devise a decay series for `""_99^240"Es"`.

1. Map a possible decay series

a. Highlight the isotopes that represent the Belt of Stability

Here is a chart I created in Excel.

It shows in yellow the isotopes of the elements from lead to einsteinium, as listed in ptable.org.

b. Mark the box representing the starting nuclide

`""_99^240"Es"` is the blue box at the upper right (`99 p, 141 n`).

c. Mark each isotope in your decay series

The blue boxes represent a decay series that includes as many as possible of the isotopes in the Belt of Stability and ends with

`""_82^208"Pb"` (`82 p, 126 n`).

d. Label the type of radiation emitted at each step

The particles emitted are

`α, β^"+", β^"+", β^"+", β^"+", α, α, α, β^"-", β^"-", β^"-", α, α, α, α`

2. Report the number of `α, β^"+"`, and `β^"-"` particles

The decay chain involves the emission of eight α particles, four `β^"+"` particles, and three `β^"-"` particles.

`""_99^240"Es" → ""_82^208"Pb" + 8color(white)(l)""_2^4"He" + 4color(white)(l)_1^0"e" + 3color(white)(l)_text(-1)^0"e"`

Note: I much prefer an alternate pathway involving the red blocks in the diagram.

It involves the emission of eight α particles and one `β^"+"` particle.

`""_99^240"Es" → ""_82^208"Pb" + 8color(white)(l)""_2^4"He" + ""_1^0"e"`