Answer
$6.0\;\rm mm$
Work Step by Step
Let's have a look at the telescope figure in your textbook.
From the geometry of the figure 24.14 [in your textbook] using small angle approximation,
$$\tan\theta_{\rm object}\approx\theta_{\rm object}=\dfrac{h'}{f_{\rm object}}$$
and
$$\tan\theta_{\rm eye}\approx\theta_{\rm eye}=\dfrac{h'}{f_{\rm eye}}$$
where the angular magnification of a telescope is given by
$$M=-20=\dfrac{\theta_{\rm eye}}{\theta_{\rm object}}=-\dfrac{f_{\rm object}}{f_{\rm eye}}$$
And we are given that $M=-20$, so
$$ -20=-\dfrac{f_{\rm object}}{f_{\rm eye}}$$
Hence,
$$f_{\rm object}=20f_{\rm eye}\tag 1$$
We can see from the geometry of the figure 24.14 that
$$\dfrac{D_{\rm object}}{D_{\rm eye}}=\dfrac{f_{\rm object}}{f_{\rm eye}}$$
Plugging from (1);
$$\dfrac{D_{\rm object}}{D_{\rm eye}}=\dfrac{20f_{\rm eye}}{f_{\rm eye}}=20$$
Hence,
$$D_{\rm eye}=\dfrac{D_{\rm object}}{20}=\dfrac{12}{20}$$
$$D_{\rm eye}=\color{red}{\bf 0.60}\;\rm cm$$
