Answer
$107^\circ \rm C$
Work Step by Step
We know, for an ideal gas, that
$$PV=nRT$$
So, the temperature of the gas is given by
$$T=\dfrac{PV}{nR}\tag 1$$
Now we need to find the number of moles of oxygen which is given by
$$n=\dfrac{m}{M}=\dfrac{m}{2M_O}$$
where $m$ is the mass of the oxygen inside the cylinder and $M_O$ is the atomic mass of the oxygen element. Noting that the oxygen gas consists of two oxygen atoms hence the 2 in front of $M_O$.
Plugging the known;
$$n=\dfrac{100\times 10^{-3}\;\rm g}{2(16\;\rm g/mol)}$$
$$n=\bf 3.125\times 10^{-3}\;\rm mol\tag 2$$
Now we know that the pressure inside the cylinder is given by
$$P=P_{out}-P_{pull}$$
$$P=P_{out}-\dfrac{F_{pull}}{A}=P_{out}-\dfrac{F_{pull}}{\pi r^2}$$
Plugging the known;
$$P=(100\times 10^3)-\dfrac{(184)}{\pi (3\times 10^{-2})^2}$$
$$P=\bf 3.49\times 10^4\;\rm Pa\tag 3$$
The volume of the gas is the volume of the cylinder which is given by
$$V=Ah=\pi r^2 h$$
Plugging the known;
$$V =\pi (3\times 10^{-2})^2 (10\times 10^{-2})$$
$$V=\bf 2.83\times 10^{-4}\;\rm m^3\tag 4$$
Plugging the known and from (2), (3), and (4) into (1);
$$T=\dfrac{(3.49\times 10^4)(2.83\times 10^{-4})}{(3.125\times 10^{-3})(8.31)}\approx \bf 380\;\rm K$$
Hence,
$$T=380-273=\color{red}{\bf 107^\circ}\rm C$$