Answer
$24.3$ cm to the left from the lens.
Work Step by Step
First, we need to find the focal distance of the lens, which is given by the +5 D lens which is a reciprocal of focal length in SI units.
Thus, $5=\dfrac{1}{f}$, and hence, $f=1/5=0.2$ m.
Hence, the focal distance of the given lens (we call it 1) is
$$f_1=20\;\rm cm\tag 1$$
We know that the image of this lens is the object of the mirror, so we need to find this image length.
$$\dfrac{1}{f_1}=\dfrac{1}{d_{o1}}+\dfrac{1}{d_{i1}}$$
Solving for $d_{i1}$;
$$d_{i1}=\left[\dfrac{1}{f_1}-\dfrac{1}{d_{o1}} \right]^{-1}$$
Plugging the known;
$$d_{i1}=\left[\dfrac{1}{20}-\dfrac{1}{30} \right]^{-1}=\bf60 \;\rm cm$$
This means that the object of the mirror is 15 cm in front of it since the distance between the lens and the mirror is 75; $d_{o2}=15\;\rm cm$.
Now we need to find the image length from the mirror.
$$\dfrac{1}{f_2}=\dfrac{1}{d_{o2}}+\dfrac{1}{d_{i2}}$$
Solving for $d_{i2}$;
$$d_{i2}=\left[\dfrac{1}{f_2}-\dfrac{1}{d_{o2}} \right]^{-1}$$
Plugging the known;
$$d_{i2}=\left[\dfrac{1}{25}-\dfrac{1}{15} \right]^{-1}=\bf -37.5\;\rm cm$$
It is obvious that this image is virtual and behind the mirror. And since the mirror reflects light back through the lens, this image is a new object to the lens, and its distance to the lens is given by $d'_{o1}=75+37.5=\bf 112.5\;\rm cm$.
Now we need to find the position of the final image.
$$\dfrac{1}{f_1}=\dfrac{1}{d_{o1}'}+\dfrac{1}{d_{i1}'}$$
Solving for $d_{i1}$;
$$d_{i1}'=\left[\dfrac{1}{f_1}-\dfrac{1}{d_{o1}'} \right]^{-1}$$
Plugging the known;
$$d_{i1}'=\left[\dfrac{1}{20}-\dfrac{1}{112.5} \right]^{-1}=\color{red}{\bf 24.3}\;\rm cm$$
This means that the final image is on the same side as the original object 24.3 cm away from the lens.
