Answer (C): There are two possible positions for the third pipe—either nestled in the gap between the pipes or outside. See the figure below.

Consider the blown-up figure below. In this diagram, $A$ is the center of the circle of radius $1$, $B$ is the center of the circle of radius $\frac{1}{4}$, and $C$ is the center of the third circle nestled in the gap. The horizontal lines through $B$ and $C$ intersect the vertical line through $A$ at $D$ and $E$, respectively, and $F$ is the foot of the perpendicular from $C$ to $BD$.

Because $AB = 1 + \frac{1}{4} = \frac{5}{4}$ and $AD = 1 - \frac{1}{4} = \frac{3}{4}$, it follows that $\triangle ADB$ is a $3$-$4$-$5$ right triangle scaled down by a factor of $4$, so $BD = \frac{4}{4} = 1$. Thus the vertical line through $B$ is tangent to the given circle of radius $1$. Then by symmetry, the radius of the larger of the two dashed circles tangent to both given circles has radius $1$.

It remains to compute the radius $r$ of the smaller dashed tangent circle. Let $x = CE = DF$. The Pythagorean Theorem in $\triangle AEC$ gives $x^2 + (1-r)^2 = (1+r)^2$, which simplifies to $x^2 = 4r$. The Pythagorean Theorem in $\triangle BFC$ gives
$$(1-x)^2 + \left(1 - \frac{3}{4} - r\right)^2 = \left(\frac{1}{4} + r\right)^2,$$
which simplifies to $(1-x)^2 = r$. Combining these equations gives $x^2 = 4(1-x)^2$, which is equivalent to $3x^2 - 8x + 4 = 0$ and $(3x-2)(x-2) = 0$. Because $x < 1$, the relevant solution is $x = \frac{2}{3}$, and the radius of the smaller circle is $\frac{1}{4} \cdot (\frac{2}{3})^2 = \frac{1}{9}$.

The requested sum of possible radii is $1 + \frac{1}{9} = \frac{10}{9}$.