= Solution
Use the local shock rest frame. Let $u$ and $d$ label upstream and downstream gas, with positive normal speeds $v_u,v_d$ through the shock. The unshocked ISM is almost stationary in the stellar frame, so $v_u\simeq V_s$, whereas the shocked gas moves outward at $V_s-v_d$ in that frame. For an ideal monatomic gas, $\gamma=5/3$; neglect upstream <pressure> and enthalpy in the strong-shock limit. Conservation of <mass>, momentum and <energy> per unit <mass> gives
$$
\rho_uv_u=\rho_dv_d,\qquad p_d+\rho_dv_d^2=\rho_uv_u^2,\qquad
\frac{v_u^2}{2}=\frac{v_d^2}{2}+\frac\gamma{\gamma-1}\frac{p_d}{\rho_d}.
$$
Put $q=\rho_d/\rho_u=v_u/v_d$. The momentum equation gives $p_d=\rho_uv_u^2(1-1/q)$. Substitution into the <energy> equation gives
$$
1=\frac1{q^2}+\frac{2\gamma}{\gamma-1}\frac{q-1}{q^2}.
$$
Discarding the unshocked root $q=1$ leaves
$$
\boxed{q=\frac{\gamma+1}{\gamma-1}=4,\quad v_d=\frac{V_s}{4},\quad
p_d=\frac34\rho_uV_s^2}.
$$
The downstream speed in the stellar frame is therefore $3V_s/4$, not $V_s/4$. These conditions describe immediate shock heating; subsequent cooling and evolution behind the front require the downstream flow profile. Cosmic-ray <pressure>, radiative <energy> loss inside the shock layer or a different equation of state would alter the compression ratio.
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