Answer
$V_3 = 0.28 m^3$
Work Step by Step
Since the process $2 \rightarrow 3 $ is adiabatic, this means the volume and temperature before and after process remain constant with $\gamma = \frac{5}{2}$
$T_2V_2^{\gamma -1} = T_3V_3^{\gamma - 1}$
$(350K)(0.226 m^3)^{2/3} = (300K)V_3^{2/3}$
$(\frac{V_3}{0.226 m^3})^{2/3} = \frac{350K}{300K} $
$\frac{V_3}{0.226 m^3}= (\frac{350K}{300K}) ^{3/2} $
$V_3= 1.26 (0.226 m^3)$
$V_3 = 0.28 m^3$
