The length of a magnet is large compared to its width and breadth....

The time period of oscillations of magnet

$T = 2\pi\sqrt{\left(\dfrac{I}{M{B}_{H}}\right)} \ldots \left(i\right)$

Where $I =$ moment of inertia of magnet

$= \dfrac{m{L}^{2}}{12}$ (m, being the mass of magnet)

When the three equal parts of magnet are placed on one another with their like poles together, then

${I}^{′} = \dfrac{1}{12}\left(\dfrac{m}{3}\right) \times {\left(\dfrac{L}{3}\right)}^{2} \times 3$

$= \dfrac{1}{12}\dfrac{m{L}^{2}}{9} = \dfrac{I}{9}$

and ${M}^{′} =$ pole strength $\times \dfrac{L}{3} \times 3 = M$

Hence,

$⇒ {T}^{′} = \dfrac{1}{3} \times T$

$⇒ {T}^{′} = \dfrac{150}{3} = 50ms$