Answer
See the detailed answer below.
Work Step by Step
Since the piston rod does 400 J of work to (rapidly) compress the air, the air undergoes an adiabatic process.
We know that: $P_1=1\;\rm atm$, $V_1=600\;\rm cm^3$, and $T_1=30^\circ\rm C$, so we can find the number of moles of the air by applying the ideal gas law.
$$PV=nRT$$
Hence,
$$n=\dfrac{P_1V_1}{RT_1}=\dfrac{(1.013\times 10^5)(600\times 10^{-6})}{(8.31)(30+273)}=\bf 0.0241\;\rm mol$$
For an adiabatic process, the heat exchange $Q$ is zero. So
$$\Delta E_{th}=Q+W=W$$
Hence,
$$nC_{\rm v}(T_2-T_1)=W$$
Thus,
$$T_2=\dfrac{W}{nC_{\rm v}}+T_1$$
Plugging the known; (the air is considered as a diatomic gas that contains mostly of nitrogen, so $C_{\rm v}=20.8$)
$$T_2=\dfrac{400}{ (0.0241)(20.8)}+(30+273)=\color{red}{\bf 1101}\;\rm K$$
We know, for an adiabatic process, that
$$T_1V^{\gamma-1}_1=T_2V^{\gamma-1}_2$$
Hence,
$$\dfrac{V_2}{V_1}=\left[\dfrac{T_1}{T_2}\right]^\frac{1}{\gamma-1}$$
$$ V_2 =V_1\left[\dfrac{T_1}{T_2}\right]^\frac{1}{\gamma-1}$$
Plugging the known;
$$ V_2 =600\left[\dfrac{(30+273)}{1101}\right]^\frac{1}{1.4-1}$$
$$V_2\approx \color{red}{\bf 24}\;\rm cm^3$$
$$P_1V^{\gamma}_1=P_2V^{\gamma}_2$$
Hence,
$$P_2=P_1\left[\dfrac{V_1}{V_2}\right]^\gamma$$
Plugging the known;
$$P_2=(1)\left[\dfrac{600}{24}\right]^{1.4}=\color{red}{\bf 91}\;\rm atm$$
