A thin convex lens of focal length 20 cm and a concave mirror of radius of curvature 20 cm stand on…

Step 1. The lens (a). $\dfrac1v = \dfrac1f - \dfrac1u = \dfrac{1}{20\ \text{cm}} - \dfrac{1}{30\ \text{cm}}$, so $v = 60\ \text{cm}$, inverted, twice the size. That is $10\ \text{cm}$ behind the mirror.

Step 2. The mirror (b). The beam is converging towards a point $10\ \text{cm}$ behind the mirror: a virtual object. With distances in front of the mirror positive, $u = -10\ \text{cm}$ and $f = +10\ \text{cm}$:

A real image $5.0\ \text{cm}$ in front of the mirror. Its magnification is $-v/u = +0.5$.

Step 3. Back through the lens (c). This image is $45\ \text{cm}$ from the lens and is a real object for it: $\dfrac1v = \dfrac{1}{20\ \text{cm}} - \dfrac{1}{45\ \text{cm}}$, so $v = 36\ \text{cm}$, on the object's side of the lens, $6\ \text{cm}$ farther from the lens than the object.

Step 4. Size (d). The magnifications are $-2$, $+0.5$ and $-0.8$. Their product is $+0.8$, so the image is upright and $1.60\ \text{cm}$ tall.

Step 5. Back onto the object (e). The light returns along its own path if the lens's image falls on the mirror's centre of curvature (mirror at $60 + 20 = 80\ \text{cm}$) or on its pole (mirror at $60\ \text{cm}$). At $80\ \text{cm}$ the final image is inverted; at $60\ \text{cm}$ it is upright.

Real or virtual. The final light converges to a point in front of the lens, on the object's side, so the image is real. A screen placed there would show it, and an eye placed well beyond it, looking towards the lens, sees it hanging in the air.