d
Given: Separation of lens for two of its position, \(d=10 \mathrm{cm}\)

Ratio of size of the images in two positions

\(\frac{I_{1}}{I_{2}}=\frac{3}{2}\)

Distance of object from the screen, \(D=?\)

Applying formula,

\({\frac{{{I_1}}}{{{I_2}}} = \frac{{{{(D + d)}^2}}}{{{{(D - d)}^2}}}}\)

\({ \Rightarrow \,\,\frac{3}{2} = \frac{{{{(D + 10)}^2}}}{{{{(D - 10)}^2}}}}\)

\({ \Rightarrow \quad \frac{3}{2} = \frac{{{D^2} + 100 + 20D}}{{{D^2} + 100 - 20D}}}\)

\(\Rightarrow 3 D^{2}+300-60 D=2 D^{2}+200+40 D\)

\(\Rightarrow \quad D^{2}-100 D+100=0\)

On solving, we get \(D=99\, \mathrm{cm}\) 

Hence the distance between the screen and the object is \(99\, \mathrm{cm} .\)