Explanation

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