Answer
$d_i=-0.40m, M=0.20$
Work Step by Step
We can find the image location as
$d_i=(\frac{1}{f}-\frac{1}{d_{\circ}})^{-1}$
We plug in the known values to obtain:
$d_i=(\frac{1}{-0.50}-\frac{1}{2.0})^{-1}$
$d_i=-0.40m$
Now the magnification is given as
$M=-\frac{d_i}{d_{\circ}}$
$\implies M=-\frac{1}{2.0}(\frac{1}{-0.50}-\frac{1}{2.0})^{-1}$
$M=0.20$