1. Let the number of $\alpha$ particles emitted be $a$ and the number of $\beta^{-}$ particles emitted be $b$.

2. Use conservation of mass number $A$:

   • Initial nucleus: ${}_{90}^{230}\mathrm{Th}$
   • Final nucleus: ${}_{84}^{214}\mathrm{Po}$

   Each $\alpha$ decay reduces mass number by $4$, while $\beta^-$ decay does not change mass number.

   So, $230 - 4a = 214$ $4a = 16$ $a = 4$

3. Now use conservation of atomic number $Z$:

   • Initial atomic number $= 90$
   • Final atomic number $= 84$

   Each $\alpha$ decay reduces atomic number by $2$ and each $\beta^-$ decay increases atomic number by $1$.

   Therefore, $90 - 2a + b = 84$

   Substitute $a=4$: $90 - 8 + b = 84$ $82 + b = 84$ $b = 2$

4. Hence,

   • Number of $\alpha$ particles $= 4$
   • Number of $\beta^-$ particles $= 2$

   Required ratio: $\frac{\text{number of }\alpha}{\text{number of }\beta^-} = \frac{4}{2} = 2$

5. Therefore, the integer answer is: $\boxed{2}$

Comparison with stored correct answer:

• Stored correct answer = $2$
• Derived answer = $2$
• They match.