Answer
$23\;\rm cm$
Work Step by Step
The author told us that Yang can see objects 150 cm away with relaxed eyes. So if he is going to wear glasses we need a lens that will create a virtual image at $s'=150$ cm when the object is far away at a distance of $s=\infty$.
Hence, the focal length of this lens is given by the thin lens equation,
$$\dfrac{1}{f }=\dfrac{1}{s_1}+\dfrac{1}{s_1'}$$
Thus,
$$f=\left[\dfrac{1}{s_1 }+\dfrac{1}{s_1'}\right]^{-1}$$
$$f=\left[\dfrac{1}{\infty}+\dfrac{1}{-150}\right]^{-1}=\bf -150\;\rm cm$$
Now we need to find the near point of his eyes when the object is at a distance $s_2$ away from his eyes while the image is then virtual and at -20 cm.
$$\dfrac{1}{f }=\dfrac{1}{s_2}+\dfrac{1}{s_2'}$$
Thus,
$$s_2=\text{Near point}=\left[ \dfrac{1}{f }-\dfrac{1}{s_2'} \right]^{-1}$$
$$ \text{Near point}=\left[ \dfrac{1}{-150}-\dfrac{1}{-20} \right]^{-1}$$
$$ \text{Near point}=\color{red}{\bf 23.1}\;\rm cm$$
