Two thin bar magnets of equal mass and length have magnetic moments...

Let $I$ be the moment of inertia of each magnet. The total moment of inertia of the combination is $I_{eq} = I + I = 2I$ in both cases.

The time period of oscillation is given by $T = 2\pi\sqrt{\dfrac{I_{eq}}{M_{eq}B}}$

When like poles are on the same side (sum configuration), the equivalent magnetic moment is $M_{eq} = M_1 + M_2$.
Thus, $T_1 = 2\pi\sqrt{\dfrac{2I}{(M_1 + M_2)B}} = 3\text{ s}$

When opposite poles are on the same side (difference configuration), the equivalent magnetic moment is $M_{eq} = M_1 - M_2$.
Thus, $T_2 = 2\pi\sqrt{\dfrac{2I}{(M_1 - M_2)B}} = 6\text{ s}$

Taking the ratio of the two time periods:
$\dfrac{T_1}{T_2} = \sqrt{\dfrac{M_1 - M_2}{M_1 + M_2}}$
$\dfrac{3}{6} = \sqrt{\dfrac{M_1 - M_2}{M_1 + M_2}}$
$\dfrac{1}{2} = \sqrt{\dfrac{M_1 - M_2}{M_1 + M_2}}$

Squaring both sides:
$\dfrac{1}{4} = \dfrac{M_1 - M_2}{M_1 + M_2}$
$M_1 + M_2 = 4M_1 - 4M_2$
$5M_2 = 3M_1 \Rightarrow \dfrac{M_1}{M_2} = \dfrac{5}{3}$

Answer: $5:3$