A thin convex lens of focal length 20 cm has a small object placed 30 cm in front of it on its principal axis. A plane mirror is placed perpendicular…

First refraction: $\dfrac1v=\dfrac1f+\dfrac1u=\dfrac1{20}-\dfrac1{30}$ gives $v=60\ \mathrm{cm}$ with magnification $m_1=v/u=-2$ (inverted, twice the size). The mirror at $50\ \mathrm{cm}$ intercepts the converging beam before it comes to a focus, so $I_1$ acts as a virtual object $10\ \mathrm{cm}$ behind the mirror. A plane mirror forms a real image $10\ \mathrm{cm}$ in front of it, i.e. $40\ \mathrm{cm}$ from the lens, with $m_2=+1$. For the returning light this point is a real object at $u=40\ \mathrm{cm}$, so $v=\dfrac{uf}{u-f}=40\ \mathrm{cm}$ on the object side, with $m_3=-v/u=-1$. The net magnification is $(-2)(+1)(-1)=+2$: real, upright relative to the original object, twice as large. The second option forgets the inversion by the second pass. The fourth would hold only if the mirror were at $40\ \mathrm{cm}$.