Answer
$\approx 7\bf\times$
Work Step by Step
First, we need to find the focal length of the eyepiece by using the mentioned equation of 25-3, as the author told us.
$$M= \dfrac{f_o}{f_e}$$
Thus,
$$f_e=\dfrac{f_o}{M}$$
Plugging the given;
$$f_e=\dfrac{26}{6.5}=\bf 4\;\rm cm$$
Now we need to use the following formula to find the image distance,
$$\dfrac{1}{f_o}=\dfrac{1}{d_{i1}}+\dfrac{1}{d_{o1}}$$
Solving for $d_{i1}$;
$$d_{i1}=\left[\dfrac{1}{f_o}-\dfrac{1}{d_{o1}}\right]^{-1}$$
Plugging the known;
$$d_{i1}=\left[\dfrac{1}{26}-\dfrac{1}{400}\right]^{-1}=\bf 27.8\;\rm cm$$
Now we can calculate the angular magnification which is given by
$$M= \dfrac{\theta'}{\theta}$$
And from the figure below;
$$M=\dfrac{\theta'}{\theta}=\dfrac{h/f_e}{h/d_{i1}}=\dfrac{d_{i1}}{f_e}$$
Plugging the known;
$$M =\dfrac{d_{i1}}{f_e}=\dfrac{27.8}{4}=\color{red}{\bf 6.95\bf \times }$$