Answer
$5\;\rm cm$
Work Step by Step
We know that the lens maker’s equation is
$$\dfrac{1}{f}=(n-1)\left[ \dfrac{1}{R_1}-\dfrac{1}{R_2} \right]$$
And since the lens is symmetric, $R_1=R_2$ where $R_1=R$, and $R_2=-R$.
Thus,
$$\dfrac{1}{f}=(n-1)\left[ \dfrac{1}{R }+\dfrac{1}{R } \right]=\dfrac{2(n-1)}{R}$$
$$f=\dfrac{R}{2(n-1)}\tag 1$$
We also know that the magnification of a telescope is given by
$$M=-\dfrac{f_{\rm obj}}{f_{\rm eye}}=20$$
Hence, in magnitudes,
$$f_{\rm obj}=20f_{\rm eye}\tag 2$$
Now we need to find the radius of curvature of the eye lens. So we need to solve (1) for $R$,
$$R_{\rm eye}=2f_{\rm eye}(n-1)$$
where, from (2), $f_{\rm eye}=f_{\rm obj}/20$
$$R_{\rm eye}=\dfrac{2f_{\rm obj}(n-1)}{20}\tag 3$$
Using (1) again to find $f_{\rm obj}$,
$$f_{\rm obj}=\dfrac{R_{\rm obj}}{2(n-1)} $$
Plug into (3),
$$R_{\rm eye}=\dfrac{R_{\rm obj}}{2(n-1)}\dfrac{2 (n-1)}{20} =\dfrac{R_{\rm obj}}{20}$$
$$R_{\rm eye} =\dfrac{100}{20}=\color{red}{\bf 5.0}\;\rm cm$$
