Answer
$13.3\;\rm cm$
Work Step by Step
We know, for thin lenses, that
$$\dfrac{1}{s}+\dfrac{1}{s'}=\dfrac{1}{f}$$
where $f=R/2$
$$\dfrac{1}{s}+\dfrac{1}{s'}=\dfrac{2}{R}\tag 1$$
We are given that the image is 3 times the object and an upright one, so
$$m=-\dfrac{s'}{s}=3$$
Hence,
$$s'=-3s$$
Plugging into (1),
$$\dfrac{1}{s}+\dfrac{1}{-3s}=\dfrac{2}{R} $$
$$ \dfrac{-3+1}{-3s}=\dfrac{2}{3s}=\dfrac{2}{R} $$
Thus,
$$s=\dfrac{R}{3}=\dfrac{40}{3}$$
$$s=\color{red}{\bf 13.3}\;\rm cm$$
