detailed solution

Correct option is A

$\begin{array}{l}\mathrm{i}=10\mathrm{A}\cdot \left(\widehat{\mathrm{K}}\right)\\ \mathrm{B}=2\times {10}^{-3}\text{ T South to }\mathrm{North}\left(\widehat{\mathrm{J}}\right)\end{array}$
To cancel the magnetic field the point should be chosen so that the net magnetic field is along - $\widehat{\mathrm{J}}$  direction
$\therefore$The point is along -$\widehat{\mathrm{i}}$  direction or along west of the wire $\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\pi r}$
$\Rightarrow 2\times {10}^{-3}=\frac{4\mathrm{\pi }\times {10}^{-7}\times 10}{2\mathrm{\pi }\times \mathrm{r}}\Rightarrow \mathrm{r}=\frac{2\times {10}^{-7}\times 10}{2\times {10}^{-3}}={10}^{-3}\mathrm{m}=1\mathrm{mm}$

detailed solution

Correct answer is 1

$\begin{array}{l}\mathrm{i}=10\mathrm{A}\cdot \left(\widehat{\mathrm{K}}\right)\\ \mathrm{B}=2\times {10}^{-3}\text{ T South to }\mathrm{North}\left(\widehat{\mathrm{J}}\right)\end{array}$
To cancel the magnetic field the point should be chosen so that the net magnetic field is along - $\widehat{\mathrm{J}}$  direction
$\therefore$The point is along -$\widehat{\mathrm{i}}$  direction or along west of the wire $\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{i}}{2\pi r}$
$\Rightarrow 2\times {10}^{-3}=\frac{4\mathrm{\pi }\times {10}^{-7}\times 10}{2\mathrm{\pi }\times \mathrm{r}}\Rightarrow \mathrm{r}=\frac{2\times {10}^{-7}\times 10}{2\times {10}^{-3}}={10}^{-3}\mathrm{m}=1\mathrm{mm}$