Consider an imaginary ion ${}_{22}{}^{48}\mathrm{X}^{3-}$. The nucleus contains '$\mathrm{a}$'$\%$ more neutrons than the number of electrons in the ion. The value of '$\mathrm{a}$' is

${{}_{22}{}^{48}\mathrm{X}}^{3-}$

In an atomic representation ${}_{\mathrm{Z}}{}^{\mathrm{A}}\mathrm{X}$

Z is atomic number = 22

A is atomic mass number = 48.

No. of protons $= \mathrm{Z} = 22$

No. of neutrons $=A-Z = 48-22 = 26$

No. of electrons $=\mathrm{P}-\mathrm{charge} = 22+3 = 25$

$\%$ of extra neutrons than electrons $=\frac{26-25}{25}\times 100=4 \% = \mathrm{a}\%$

So, the value of a=4.