Answer
$$\frac{I_E}{I_M}=7.92\times10^{-4}$$
Work Step by Step
At first, only the engine has rotational motion, so $$L_0=I_E\omega=7700I_E$$
After that, the engine's speed rises to $12500rev/min$. There appears the rotation of the rest of the motorcycle, with $\omega=-3.8rev/min$ and $I_M$. Therefore, $$L=12500I_E-3.8I_M$$
Angular momentum is conserved, so $$7700I_E=12500I_E-3.8I_M$$ $$3.8I_M=4800I_E$$ $$\frac{I_E}{I_M}=7.92\times10^{-4}$$
