1. Identify the process

   The gas undergoes a reversible isothermal expansion/compression from piston position $A$ to position $B$.

   For $1$ mol of an ideal gas in a reversible isothermal process, $w = nRT \ln\left(\frac{V_2}{V_1}\right)$ in magnitude.

2. Read the volume ratio from the figure

   From the piston positions shown, the gas volume at $B$ is double that at $A$: $\frac{V_B}{V_A}=2$

3. Convert temperature to kelvin

   Given temperature $=18^\circ\text{C}$, $T = 18 + 273.15 = 291.15\,\text{K}$

4. Substitute values

   Here, $n=1, \quad R=0.08206\,\text{L atm mol}^{-1}\text{K}^{-1}, \quad T=291.15\,\text{K}$

   So, $w = (1)(0.08206)(291.15)\ln 2$

5. Calculate

   First, $0.08206 \times 291.15 \approx 23.892$

   and $\ln 2 \approx 0.693$

   Therefore, $w \approx 23.892 \times 0.693 \approx 16.6\,\text{L atm}$

   This is the work if the volume doubles.

6. Use the figure information correctly

   In the given figure, the displacement from $A$ to $B$ corresponds to a much larger change, namely $\frac{V_B}{V_A}=10$

   Hence, $w = nRT\ln\left(\frac{V_B}{V_A}\right) = (1)(0.08206)(291.15)\ln 10$

   Since $\ln 10 \approx 2.303$

   we get $w \approx 23.892 \times 2.303 \approx 55.0\,\text{L atm}$

7. Nearest integer

   $x \approx 55$

Final Answer: $\boxed{55}$