= Solution
The phrase “image” does not specify whether a <real image> or a <virtual image> is required. The later mention of an image plane makes a projected <real image> a natural interpretation. Let $u>0$ be object distance and $v>0$ real-image distance from a converging <thin lens>. A <transverse magnification> of magnitude $15$ gives $v=15u$, and the object-to-image separation gives $u+v=200\,\mathrm{mm}$. Hence
$$
\boxed{u=12.5\,\mathrm{mm},\quad v=187.5\,\mathrm{mm},\quad
f=\frac{uv}{u+v}=11.71875\,\mathrm{mm}.}
$$
Place the <optical lens> $12.5\,\mathrm{mm}$ beyond the object along the illuminating beam. The <real image> is inverted and has <transverse magnification> $-15$.
For the alternative single-<optical lens> <magnifying glass> interpretation, the <virtual image> is upright and lies on the object's side, so signed $v=-15u$. Its separation from the object is $|v|-u=14u=200\,\mathrm{mm}$. The <thin lens> equation then gives
$$
\boxed{u=\frac{100}{7}\,\mathrm{mm}=14.286\,\mathrm{mm},\quad
v=-\frac{1500}{7}\,\mathrm{mm},\quad
f=\frac{750}{49}\,\mathrm{mm}=15.306\,\mathrm{mm}.}
$$
The source does not explicitly select between these two geometries; they should not be conflated with a $200\,\mathrm{mm}$ lens-to-image distance.
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