At point $P$ the needle oscillates $20$ times per minute. Therefore, time period $T_P=\frac{60}{20}= 3 \mathrm{s}$ and at point $Q$, $T_Q=\frac{60}{10}=6 \mathrm{s}$.

The time period of oscillation magnetometer is given by,

$T=2\pi \sqrt{\frac{I}{B_{H}M}}$

$\Rightarrow T_P=3=2\pi \sqrt{\frac{I}{\left(B_P\cos 30°\right)M}}$

And $T_Q=6 \mathrm{s}=2\pi \sqrt{\frac{I}{\left(B_Q\cos 60°\right)M}}$

Taking ratio,

$\frac{3}{6}=\sqrt{\frac{1}{\left({ B}_P\frac{\sqrt{3}}{2}\right)}\times \left(\frac{B_Q}{2}\right)}$

$\Rightarrow \frac{\sqrt{3}}{4}=\frac{B_Q}{B_P}$

$\Rightarrow B_Q:B_P=\sqrt{3}:4$