Explanation

(b)
Charge of $e^{-} = 1.6 \times 10^{- 19}$
Dipole moment of HBr = $1.6\ \times 10^{- 30}$
Inter-atomic spacing = $1\ \mathring{\mathrm{A}}$
$\ \ = 1 \times 10^{- 10}\ m$
Percentage of ionic character in HBr
$= \frac{Dipole\ moment\ of\ HBr\ \times \ 100}{inter\ spacing\ distance\ \times \ q}$
$\ = \frac{1.6 \times 10^{- 30}}{1.6 \times 10^{- 19} \times 10^{- 10}} \times 100$
$\ \ = 10^{- 30}\ \times 10^{29} \times 100$
$= 10^{- 1} \times 100$
$= 0.1 \times 100$
$= 10\%$