Answer
$$0.215
$$
Work Step by Step
Adiabatic relations $\mathrm{Eq} .19-54$ and $\mathrm{Eq}$. $19-56$ lead to
$$
p_{3}=p_{1}\left(\frac{V_{1}}{V_{3}}\right)^{\gamma}=\frac{p_{1}}{3^{44}} \Rightarrow \frac{p_{3}}{p_{1}}=\frac{1}{3^{1.4}}=0.215
$$
