Find the final temperature and volume of the gas
The process should indeed be considered irreversible, as the problem statement gives a rather explicit hint stating that the pressure was released suddenly, so that your 3rd suggestion is correct. With an irreversible expansion against a constant external pressure, the adiabatic equation $pV^\gamma = \text{const}$ is inapplicable, so one must use the first law of thermodynamics. Work is done at the expense of the loss of internal energy:

$$W = - \Delta U = nC_\mathrm{v}(T_1 - T_2) \tag{1}$$

For ideal monoatomic gas the following is applied:

$$n = \frac{p_1V_1}{RT_1} \tag{1.1}$$
$$C_\mathrm{v} = \frac{3R}{2} \tag{1.2}$$

so that (1) can be rewritten as such:

$$W = nC_\mathrm{v}(T_1 - T_2) = \frac{3p_1V_1R}{2RT_1}(T_1 - T_2) = \frac{3p_1V_1}{2} - \frac{3p_1V_1T_2}{2T_1} \tag{1.3}$$

At the same time work by gas expansion is

$$W = p_2(V_2 - V_1) = p_2V_2 - p_2V_1 \tag{2}$$

and from the ideal gas law the reduced product is

$$p_2V_2 = nRT_2 \tag{2.1}$$

so that (2) can also be rewritten:

$$W = p_2(V_2 - V_1) = nRT_2 - p_2V_1 \tag{2.2}$$

I would leave for OP the task to to equate (1.3) and (2.2) and express the final temperature $T_2$ from there:

$$\frac{3p_1V_1}{2} - \frac{3p_1V_1T_2}{2T_1} = nRT_2 - p_2V_1$$

Here I bring up the final result:

$$T_2 = \frac{3p_1V_1 + 2p_2V_1}{5p_1V_1} T_1 = \frac{3 \cdot \pu{10 atm} \cdot \pu{10 L} + 2 \cdot \pu{1 atm} \cdot \pu{10 L}}{5 \cdot \pu{10 atm} \cdot \pu{10 L}} \cdot \pu{273.15 K} = \pu{174.7 K} \tag{3}$$

Volume after expansion is also to be found using ideal gas law and (1.1):

$$V_2 = \frac{nRT_2}{p_2} = \frac{p_1V_1RT_2}{RT_1p_2} = \frac{p_1T_2}{p_2T_1}V_1 = \frac{\pu{10 atm} \cdot \pu{174.7 K}}{\pu{1 atm} \cdot \pu{273.15 K}} \cdot \pu{10 L} = \pu{64.0 L} \tag{4}$$