A biconvex lens of focal length 15 ~cm is in front of a plane...

Object is placed at distance $2 f$ from the lens. So first image $I_1$ will be formed at distance $2 f$ on other side. This image $I_1$ will behave like a virtual object for mirror. The second image $I_2$ will be formed at distance $20 \mathrm{~cm}$ in front of the mirror, or at distance $10 \mathrm{~cm}$ to the left hand side of the lens.
Now applying lens formula $$\begin{array}{rlr}& \frac{1}{v}-\frac{1}{u}=\frac{1}{f}\\ & \frac{1}{v}-\frac{1}{+10}=\frac{1}{+15}\\ \text{ or }& v=6 \mathrm{~cm} \end{array}$$Therefore, the final image is at distance $16 \mathrm{~cm}$ from the mirror. But, this image will be real. This is because ray of light is travelling from right to left. $\therefore$ The correct option is (b).