Answer
$-6689 \ J$
Work Step by Step
We know the following equation for the work it will take:
$ W_2=\frac{ln(c_2)}{ln(c_1)}(W_1)$
Where $c_1$ is the fraction that the volume is initially changed by and where $c_2$ is the factor by which the volume will be changed by. Thus, we find:
$ W_2=\frac{ln(\frac{1}{22})}{ln(\frac{1}{2})}(-1500)=-6689 \ J$
