Answer: 173

Solution:

A sphere of radius $r$ resting in a circular hole of radius $a$ has its center on the axis of the hole; since the center is at distance $r$ from every point of the hole's rim, it sits at distance $\sqrt{r^2 - a^2}$ from the plane of the board. So the two centers lie at depths $\sqrt{r^2 - 1}$ and $\sqrt{r^2 - 4}$ on the same side of the board, with horizontal separation $7.$

Tangency of the spheres means the centers are $2r$ apart: $$\begin{aligned} &49 + \left(\sqrt{r^2 - 1} - \sqrt{r^2 - 4}\right)^2 \\ &= 4r^2. \end{aligned}$$ Expanding gives $49 + 2r^2 - 5$ $- 2\sqrt{(r^2 - 1)(r^2 - 4)}$ $= 4r^2,$ so $\sqrt{(r^2 - 1)(r^2 - 4)} = 22 - r^2.$ Squaring, $r^4 - 5r^2 + 4 = 484 - 44r^2 + r^4,$ hence $39r^2 = 480$ and $r^2 = \frac{160}{13}.$

Thus $m + n = 160 + 13 = 173.$