${v_o} = \frac{{{u_o}{f_o}}}{{{u_o} - {f_o}}} = \frac{{3 \times 2}}{{3 - 2}} = 6\,cm$
Now as in case of lenses in contact
$\frac{1}{{{F_o}}} = \frac{1}{{{f_1}}} + \frac{1}{{{f_2}}} + \frac{1}{{{f_3}}} + ..... = \frac{1}{{{f_1}}} + \frac{1}{{{{F'}_o}}}$
where, $ \frac{1}{{{{F'}_o}}} = \frac{1}{{{f_2}}} + \frac{1}{{{f_3}}} + ...... $
So if one of the lens is removed, the focal length of the remaining lens system
$\frac{1}{{{{F'}_o}}} = \frac{1}{{{F_0}}} - \frac{1}{{{f_1}}} = \frac{1}{2} - \frac{1}{{10}}$ ==> ${F'_o} = 2.5\,cm$
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
$i.e. \,15 -6 = 9\, cm.$
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