Answer
$\frac{T_4}{T_1} = 0.660$
Work Step by Step
Step $4 \rightarrow 1$ is adiabatic which means there is no heat transfer in the process.
$T_4V_4 ^{\gamma -1} = T_1V_1 ^{\gamma -1} $
Since $V_4 = 4.00V_1$,
and $\gamma = 1.30$
$\frac{T_4}{T_1} = [\frac{V_1}{V_4}]^{\gamma -1} $
$\frac{T_4}{T_1} = [\frac{1.00}{4.00 }]^{0.30} $
$\frac{T_4}{T_1} = 0.660$
