$\mathrm{B}_{\text {coil }}=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{r}} \quad \text { (outwards) }$

$B$ due to wire,

$\mathrm{B}_{\text {wire }}=\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}$

${\text { Given }: \mathrm{i}=8\, \mathrm{A}, \mathrm{r}=10 \times 10^{-2} \,\mathrm{m}}$

$\frac{{{\mu _0}}}{{4\pi }} = {10^{ - 7}}$

Magnetic field at centre $\mathrm{C}$

$\mathrm{B}_{\mathrm{C}}=\mathrm{B}_{\text {coil }}+\mathrm{B}_{\text {wire }}$

${=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{r}}(\mathrm{upward})+\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}}$ (inwards)

${=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{r}}-\frac{\mu_{0} \mathrm{i}}{2 \pi \mathrm{r}}=\frac{\mu_{0} \mathrm{i}}{2 \mathrm{r}}\left(1-\frac{1}{\pi}\right)}$ (outwards)

$=\frac{4 \pi \times 10^{-7} \times 8}{2 \times 10 \times 10^{-2}}\left(1-\frac{1}{3.14}\right) \quad$ (outwards)

$=\frac{4 \times 3.14 \times 10^{-7} \times 8 \times 2.14}{2 \times 10 \times 10^{-2} \times 3.14} \quad$ (outwards)

$=3.424 \times 10^{-5} \quad$ (outwards)