Answer
$i=-3.8cm$
Work Step by Step
Since the lens is diverging, the focal length is negative (f=-6.0cm). Use the lens equation $$\frac{1}{f}=\frac{1}{i}+\frac{1}{p}$$ to solve for $i$. $$\frac{1}{i}=\frac{p}{pf}-\frac{f}{pf}$$ $$\frac{1}{i}=\frac{p-f}{pf}$$ $$i=\frac{pf}{p-f}$$ Substituting values of $f=-6.0cm$ and $p=10.cm$ yields an image distance of $$i=\frac{(10.cm)(-6.0cm)}{6.0cm+10.cm}=-3.8cm$$
