Answer
The ratio of KE of the hydrogen atom after the collision to that of the electron before the collision is $2.2\times10^{-3}$
Work Step by Step
The electron, whose mass is $m$ and which has initial velocity $\vec{v_0}$, hits (elastically) the hydrogen atom, whose mass is $M$ and which has initial velocity $\vec{V_0}=0$.
We assume the total linear momentum before and after the collision is conserved, so $$M\vec{V_0}+m\vec{v_0}=M\vec{V}_f+m\vec{v}_f$$ $$mv_0=MV_f+mv_f$$ $$v_f=v_0-\frac{M}{m}V_f (1)$$
Also, since the collision is elastic, the kinetic energy before and after the collision is conserved: $$\frac{1}{2}(MV^2_0+mv_0^2)=\frac{1}{2}(MV_f^2+mv_f^2)$$ $$mv_0^2=MV_f^2+mv_f^2$$ $$v_f^2=v_0^2-\frac{M}{m}V_f^2 (2)$$
Taking (1) to the power of 2 and setting it equal to (2), we have $$v_0^2+\frac{M^2}{m^2}V_f^2-2\frac{M}{m}V_fv_0=v_0^2-\frac{M}{m}V_f^2$$
Eliminate $v_0^2$ from both sides: $$\frac{M^2}{m^2}V_f^2-2\frac{M}{m}V_fv_0=-\frac{M}{m}V_f^2$$ $$V_f^2\Big(\frac{M^2}{m^2}+\frac{M}{m}\Big)=2\frac{M}{m}V_fv_0$$ $$V_f\Big(\frac{M^2}{m^2}+\frac{M}{m}\Big)=2\frac{M}{m}v_0$$ $$\frac{V_f}{v_0}=\frac{2\frac{M}{m}}{\frac{M^2}{m^2}+\frac{M}{m}}=\frac{2}{\frac{M}{m}+1}$$
The ratio of KE of the hydrogen atom after the collision to that of the electron before the collision is $$R=\frac{\frac{1}{2}MV_f^2}{\frac{1}{2}mv_0^2}=\frac{M}{m}\times\frac{V_f^2}{v_0^2}=\frac{M}{m}\times\frac{4}{\Big(\frac{M}{m}+1\Big)^2}$$
We know $\frac{M}{m}=1837$, so $$R=\frac{4\times1837}{1838^2}=2.2\times10^{-3}$$
