Answer: $\mathbf C$
Given:
$AC = 15$
$BC = 10$
$\because$ $D$ is the median.
$\therefore BD = CD = 5$
Using $\text{Cosine Rule}$
$$\cos B = \frac{a^2+c^2-b^2}{2ac}$$
$$\therefore \cos B = \frac{ AB^2 + BC^2 -2AC^2}{2AB\times BC}$$
$$\implies \cos B = \frac{49 + 100 -15^2}{140}$$
$$\implies \cos B = \frac{-19}{35}$$
Now, we have:
$$AD^2 = AB^2 + BD^2 -2AB\times BD \cos B$$
$$\implies AD^2 = 49+25-2\times7\times5\bigg(\frac{-19}{35}\bigg)$$
$$\implies AD^2 = 112$$
$$\implies AD = \sqrt{112}$$
$\therefore\;\bf{C}$ is the correct option.