Magnetic Field due to a Current Element

Magnetic field on the axis of a circular coil at distance x from centre, B = $\frac{{\mu }_{0}Ni{r}^2}{2{\left(r^2+x^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$ $\frac{{}_{}{}^{}}{{{}^{}{}^{}}^{\raisebox{1ex}{$$}\!\left/ \!\raisebox{-1ex}{$$}\right.}}$

$\frac{{\left(r^2+{\left(0.2\right)}^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\left(r^2+{\left(0.05\right)}^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=8$ $\frac{r^2+{\left(0.2\right)}^2}{r^2+{\left(0.05\right)}^2}=4\Rightarrow r^2=0.01\Rightarrow r=0.1m.$

Using magnetic field due to straight wire :

$B=\frac{{\mu }_{0}i}{4\pi r}\left(sin\alpha +sin\beta \right)$

$=\frac{10^{-7}\times 1.5}{\left(\frac{0.09}{2\sqrt[]{3}}\right)}\times \left(sin60°+sin60°\right)=10^{-5}T$

So, magnetic field due to three wires

= 3 × 10^-5 T

inside the plane