Explanation

(b) $B$ at the centre of a coil carrying a current, $i$ is
$B_{Coil} = \frac{\mu_{0}i}{2r}$ [upward]
$B$ due to wire $B_{wire} = \frac{\mu_{0}i}{2\pi r}$ [downward]
Given $i = 8A;r = 10 \times 10^{- 2}m$
$\frac{\mu_{0}}{4\pi} = 10^{- 7}$
Magnetic field at centre $C$,
$B_{C} = B_{coil} + B_{wire}$
$= \frac{\mu_{0}i}{2r}$(upward) $+ \frac{\mu_{0}i}{2\pi r}$ [downward]
$= \frac{\mu_{0}i}{2r} - \frac{\mu_{0}i}{2\pi r} = \frac{\mu_{0}i}{2r}\left( 1 - \frac{1}{\pi} \right)\ upward$
$= \frac{4\pi \times 10^{- 7} \times 8}{2 \times 10 \times 10^{- 2}}\left( 1 - \frac{1}{3.14} \right)upward$
$= \frac{4 \times 3.14 \times 10^{- 7} \times 8 \times 2.14}{2 \times 10 \times 10^{- 2} \times 3.14} = 3.424 \times 10^{- 5}\ upward$