Perfect elastic collision, one dimensional
Using conservation of momentum and energy
Final velocity of $8~\mu g$ particle: $v_1' = \dfrac{m_1 - m_2}{m_1 + m_2}v$
Final velocity of $4~\mu g$ particle: $v_2' = \dfrac{2m_1}{m_1 + m_2}v$
de Broglie wavelength $\lambda = \dfrac{h}{mv}$
So $\lambda_1 : \lambda_2 = \dfrac{1}{m_1 v_1'} : \dfrac{1}{m_2 v_2'} = \dfrac{1}{8 \cdot v_1'} : \dfrac{1}{4 \cdot v_2'} = 2:1$