#### Answer

40 J of work is done on the gas.

#### Work Step by Step

We can calculate the work done as:
$W = P\Delta V$
$W = P ~(V_f-V_i)$
$W = (2.00\times 10^5~Pa)(100\times 10^{-6}~m^3-300\times 10^{-6}~m^3)$
$W = -40~J$
The minus sign shows that 40 J of work is done on the gas.