Explanation

(d) If initially the objective (focal length $F_{o}$) forms the image at distance $v_{o}$ then $v_{o} = \frac{u_{o}f_{o}}{u_{o} - f_{o}} = \frac{3 \times 2}{3 - 2} = 6\ cm$
Now as in case of lenses in contact
$\frac{1}{F_{o}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} + \frac{1}{f_{3}} + \ldots = \frac{1}{f_{1}} + \frac{1}{F_{o}^{'}}$
$\left\{ where\frac{1}{F_{o}^{'}} = \frac{1}{f_{2}} + \frac{1}{f_{3}} + \ldots \right\}$
So if one of the lens is removed, the focal length of the remaining lens system
$\frac{1}{F_{o}^{'}} = \frac{1}{F_{0}} - \frac{1}{f_{1}} = \frac{1}{2} - \frac{1}{10} \Rightarrow F_{o}^{'} = 2.5\ cm$
This lens will form the image of same object at a distance $v_{o}'$ such that
$v_{o}^{'} = \frac{u_{o}F_{o}^{'}}{u_{o} - F_{o}^{'}} = \frac{3 \times 2.5}{(3 - 2.5)} = 15\ cm$
So to refocus the image, eye-piece must be moved by the same distance through which the image formed by the objective has shifted $i.e.\ 15 - 6 = 9\ cm$