1. Find the number of protons, neutrons, and electrons for ${}_{22}^{48}X^{3-}$.

   • Atomic number $Z = 22$ $Rightarrow$ number of protons $= 22$.
   • Mass number $A = 48$ $Rightarrow$ number of neutrons $N = A - Z = 48 - 22 = 26.$
   • Since the ion is $X^{3-}$, it has gained $3$ electrons. So number of electrons $e = 22 + 3 = 25.$

2. Interpret the statement: “The nucleus contains '$a$'% more neutrons than the number of electrons in the ion.”

   This means neutrons are $a\%$ more than electrons: $N = e\left(1 + \frac{a}{100}\right).$

3. Substitute the values: $26 = 25\left(1 + \frac{a}{100}\right).$

4. Solve for $a$: $\frac{26}{25} = 1 + \frac{a}{100}$ $1.04 = 1 + \frac{a}{100}$ $\frac{a}{100} = 0.04$ $a = 4.$

5. Nearest integer: $a = 4.$

Therefore, the required value is 4.