Answer
The mass of the asteroid should be $\frac{M}{10}$.
Work Step by Step
We can find $\omega_2$ in terms of $\omega_1$ when the length of the day increases by 25%.
$T_2 = 1.25~T_1$
$\frac{2\pi}{\omega_2} = 1.25\times \frac{2\pi}{\omega_1}$
$\omega_2 = 0.80~\omega_1$
We can use conservation of angular momentum to solve this question. Let $M$ be the mass of the earth and let $m$ be the mass of the asteroid.
$L_2 = L_1$
$I_2\omega_2 = I_1\omega_1$
$(\frac{2}{5}MR^2+mR^2)(0.80~\omega_1) = \frac{2}{5}MR^2\omega_1$
$(M+\frac{5}{2}m)(0.80)= M$
$2.0~m = 0.20~M$
$m = \frac{M}{10}$
The mass of the asteroid should be $\frac{M}{10}$.
