Formula
Field due to each coil
Magnitude of resultant field
Direction of resultant field
Field at the centre of a circular coil $=\frac{{\mathit{\mu }}_0\mathit{I}}{2\mathit{R}}$
Field due to coil $\mathit{P}=\frac{{\mathit{\mu }}_0\times 3}{2\times 5\times {10}^{-2}}$ tesla

Field due to coil $\mathrm{Q}=\frac{{\mathit{\mu }}_0\times 3}{2\times 5\times {10}^{-2}}$ tesla

$\therefore$ Resultant Field $=\left(\mathit{\pi }\sqrt{{12}^2+{16}^2}\right)\mathit{\mu }\mathit{T}$
$=\left(20\mathit{\pi }\right)\mathit{\mu }\mathrm{T}$
Let the field make an angle $\mathit{\theta }$ with the vertical
$\tan \mathit{\theta }=\frac{12\mathit{\pi }\times {10}^{-6}}{16\mathit{\pi }\times {10}^{-6}}=\frac{3}{4}$
$\mathit{\theta }={\tan }^{-1}\frac{3}{4}$