As a convex lens alone can form a real images as well as a virtual images, therefore, the lens in the present question is a convex lens. Let $f$ be the focal length of the lens and $m$ be the magnification produced

In the first case, when image is real,

$u=-16$cm,$v=(m\times 16)$cm

As $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\therefore \frac{1}{16m}+\frac{1}{16}=\frac{1}{f}$or$1+\frac{1}{m}=\frac{16}{f}$...(i)

In the second case, when image is virtual

$u=-6 \mathrm{c}\mathrm{m},v=(-6$m)cm

From $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\frac{1}{-6m}+\frac{1}{6}=\frac{1}{f}$or$1-\frac{1}{m}=\frac{6}{f}$, ...(ii)

Add Eq.(i) and Eq.(ii) we have

$2=\frac{22}{f}$or$f=\frac{22}{2}=11$cm