Answer
5 cm to the right from the lens.
Work Step by Step
The parallel rays mean that the position of the object is $s_1=\infty$.
Thus by the thin lens law
$$\dfrac{1}{f}=\dfrac{1}{s}+\dfrac{1}{s'}$$
Hence,
$$\dfrac{1}{f_1}=\dfrac{1}{\infty}+\dfrac{1}{s_1'}=0+\dfrac{1}{s_1'}$$
$$f_1=s_1' $$
This means that the image created by the given diverging lens is just at its focal point.
We are given that $f_1=-10$ cm, so that
$$s_1'=-10\;\rm cm$$
This image is on the same side as the parallel rays and this means that the distance between this image and the mirror is 30 cm.
(The first image is the object of the mirror).
Hence,
$$\dfrac{1}{f_2}=\dfrac{1}{s_2}+\dfrac{1}{s_2'}$$
Solving for $s_2'$,
$$s_2'=\left[ \dfrac{1}{f_2}-\dfrac{1}{s_2} \right]^{-1}$$
$$s_2'=\left[ \dfrac{1}{10}-\dfrac{1}{30} \right]^{-1}=\bf 15\;\rm cm$$
Thus the final image (which is a point at which the rays are focused at) is 15 cm to the left from the mirror or 5 cm to the right from the lens.
