Answer
a) $1.9\times 10^{-3} \;\rm m^3$
b) $74^\circ\;\rm C$
Work Step by Step
a) Let's assume that Oxygen gas is an ideal gas.
$$PV=nRT$$
Hence, the initial volume of the given gas is
$$V_1=\dfrac{nRT_1}{P_1}\tag 1$$
And we know, for an adiabatic process that
$$P_1V_1^\gamma=P_2 V_2^\gamma $$
Thus, the final volume is given by
$$\dfrac{V_2^\gamma}{V_1^\gamma}=\dfrac{P_1}{P_2 }$$
$$\left[ \dfrac{V_2 }{V_1}\right]^\gamma=\dfrac{P_1}{P_2 } $$
$$ V_2 =V_1\left[\dfrac{P_1}{P_2 } \right]^\frac{1}{\gamma}$$
We know, for a diatomic gas that $\gamma=1.4$,
$$ V_2 =V_1\left[\dfrac{P_1}{P_2 } \right]^\frac{1}{1.4}$$
Plugging from (1);
$$ V_2 =\dfrac{nRT_1}{P_1}\left[\dfrac{P_1}{P_2 } \right]^\frac{1}{1.4}$$
And since the the pressure is halved, $P_1=2P_2$
$$ V_2 =\dfrac{nRT_1}{P_1}\left[\dfrac{2 \color{red}{\bf\not} P_2}{ \color{red}{\bf\not} P_2 } \right]^\frac{1}{1.4}$$
$$ V_2 =\left[2^{(1/1.4)}\right] \dfrac{nRT_1}{P_1}$$
Plugging the known;
$$ V_2 =\left[2^{(1/1.4)}\right]\dfrac{(0.1)(8.31)(150+273)}{(3\times 1.013\times 10^5)}$$
$$V_2 =\color{red}{\bf 1.9\times 10^{-3}}\;\rm m^3$$
---
b) Now we need to find the final temperature of the gas, so we need to use the ideal gas law of
$$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$$
Hence,
$$T_2=\dfrac{P_2V_2}{P_1V_1}{T_1}=\dfrac{P_2V_2}{P_1 }\dfrac{1}{V_1}{T_1}$$
Plug $V_1$ from (1);
$$T_2=\dfrac{P_2V_2 }{ \color{red}{\bf\not} P_1 }\dfrac{ \color{red}{\bf\not} P_1}{nR \color{red}{\bf\not} T_1}{ \color{red}{\bf\not} T_1}$$
where $P_2=\frac{1}{2}P_1$,
$$T_2=\dfrac{P_1V_2 }{2 nR}$$
Plugging the known;
$$T_2=\dfrac{(3\times 1.013\times 10^5)(1.9\times 10^{-3}) }{2 (0.10)(8.31)}=\bf 347\;\rm K$$
$$T_2=\color{red}{\bf 74}^\circ\;\rm C$$
