A compass needle of oscillation magnetometer oscillates 20 times...

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

The time period of oscillation magnetometer is given by,

$T = 2\pi\sqrt{\dfrac{I}{{B}_{H}M}}$

$⇒ {T}_{P} = 3 = 2\pi\sqrt{\dfrac{I}{\left({B}_{P}\cos 30 °\right)M}}$

And ${T}_{Q} = 6s = 2\pi\sqrt{\dfrac{I}{\left({B}_{Q}\cos 60 °\right)M}}$

Taking ratio,

$\dfrac{3}{6} = \sqrt{\dfrac{1}{\left({B}_{P}\dfrac{\sqrt{3}}{2}\right)} \times \left(\dfrac{{B}_{Q}}{2}\right)}$

$⇒ \dfrac{\sqrt{3}}{4} = \dfrac{{B}_{Q}}{{B}_{P}}$

$⇒ {B}_{Q}: {B}_{P} = \sqrt{3}: 4$