= Solution
Work in the <shock frame>, and let the positive upstream and downstream speeds be $w_0$ and $w_1$. The <Rankine-Hugoniot conditions for a perfect gas> give conservation of mass and momentum:
$$
\rho_0w_0=\rho_1w_1,
\qquad
p_0+\rho_0w_0^2=p_1+\rho_1w_1^2.
$$
Put
$$
r=\frac{\rho_1}{\rho_0},\qquad
\frac{p_1}{p_0}=1+\beta.
$$
Mass conservation gives $w_1=w_0/r$, and momentum conservation then gives
$$
w_0^2=\frac{p_0}{\rho_0}\frac{\beta r}{r-1}. \qquad (1)
$$
Applying the energy integral from part (a) on the two sides gives
$$
\frac{\gamma}{\gamma-1}\frac{p_0}{\rho_0}+\frac12w_0^2
=\frac{\gamma}{\gamma-1}\frac{p_0(1+\beta)}{\rho_0r}
+\frac12\frac{w_0^2}{r^2}. \qquad (2)
$$
Substitution of (1) into (2), followed by cancellation of $p_0/\rho_0$, yields
$$
\frac{\beta(r+1)}2
=\frac{\gamma}{\gamma-1}(1+\beta-r).
$$
Solving this linear equation for the density ratio gives
$$
\boxed{\frac{\rho_1}{\rho_0}=r
=\frac{2\gamma+(\gamma+1)\beta}
{2\gamma+(\gamma-1)\beta}}.
$$
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