Answer
$15.7\;\rm cm$
Work Step by Step
We are given that
$$s+s'=60\;\rm cm\tag 1$$
and that the magnification is 2 but since the image is real and the object is real, the image is inverted.
$$m=-2\tag 2$$
where $m=-\dfrac{s'}{s}$ and $s'$, from (1), $=60-s$, so that
$$-2=-\dfrac{60-s}{s}$$
$$s=\bf 20\;\rm cm$$
And hence,
$$s'=\bf 40\;\rm cm$$
Now we can use the lens maker equation (see equation 23.27),
$$\dfrac{1}{s}+\dfrac{1}{s'}=\dfrac{1}{f}=(n-1)\left[\dfrac{1}{R_1}-\dfrac{1}{R_2}\right]$$
And for the symmetric converging lens, we know that $R_1=|R_2|$ where $R_1=R$, and $R_2=-R$
Thus,
$$\dfrac{1}{s}+\dfrac{1}{s'} =(n-1)\left[\dfrac{1}{R}-\dfrac{1}{-R}\right]$$
$$\dfrac{1}{s}+\dfrac{1}{s'} =(n-1)\left[\dfrac{2}{R} \right]$$
Hence, the radii of the system are then given by
$$R=\dfrac{2(n-1)}{\dfrac{1}{s}+\dfrac{1}{s'}} $$
Plugging the known;
$$R=\dfrac{2(1.59-1)}{\dfrac{1}{20}+\dfrac{1}{40}} =\bf 15.7\;\rm cm$$
Thus,
$$R_1=R_2=\color{red}{\bf 15.7}\;\rm cm$$
