First, we need to convert the pressure in SI units. Keeping in mind that [tex]1 atm = 1.01 \cdot 10^5 Pa[/tex]:
[tex]p=10.0 atm =1.01 \cdot 10^6 Pa[/tex]
The initial and final volume of the gas are (keeping in mind that [tex]1.0 L = 0.001 m^3[/tex]):
[tex]V_i = 20.0 L=0.020 m^3[/tex]
[tex]V_f = 2.0 L=0.002 m^3[/tex]
So, the work done on the gas by the surrounding is
[tex]W= -p \Delta V=-p(V_f-V_i)=-(1.01 \cdot 10^6 Pa)(0.002 m^3-0.020 m^3)=18180 J[/tex]
And the final positive sign means that this work corresponds to an increase in internal energy of the gas.
