Answer
$$\frac{m_1}{m_2}=0.71$$
Work Step by Step
Since the skaters are initially stationary, the initial total linear momentum is $0$, or $\sum p_0=0$
After that, they push each other and have opposite velocities. Taking the direction of the velocity of skater 1 to be the positive side, we have $$\sum p_f=m_1v_1-m_2v_2$$
According to the conservation of linear momentum, we have $$\sum p_0=\sum p_f$$ $$m_1v_1-m_2v_2=0$$ $$\frac{m_1}{m_2}=\frac{v_2}{v_1}$$
And then because of friction, they glide to a halt at a similar acceleration, $a$. Using an equation of kinematics, we have $$0^2=v_1^2+2as_1$$ $$0^2=v_2^2+2as_2$$ Therefore, $$v_1^2=-2as_1$$ $$v_2^2=-2as_2$$
Divide the lower two equations, $$\Big(\frac{v_2}{v_1}\Big)^2=\frac{s_2}{s_1}=\frac{1}{2}$$ $$\frac{m_1}{m_2}=\frac{v_2}{v_1}=\frac{\sqrt2}{2}=0.71$$
