1. Recall standard notation for an ellipse

   For an ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \qquad a>b$ where:

   • Length of major axis $=2a$
   • Length of minor axis $=2b$
   • Distance between the foci $=2c$
   • Eccentricity $e=\frac{c}{a}$
   • Also, $c^2=a^2-b^2$

2. Use the given condition

   The problem says:

   length of the minor axis is equal to one fourth of the distance between the foci.

   So, $2b=\frac{1}{4}(2c)$

   Simplify: $2b=\frac{c}{2}$ $4b=c$

3. Relate this to eccentricity

   Since $c=4b$, substitute into $c^2=a^2-b^2$

   We get: $(4b)^2=a^2-b^2$ $16b^2=a^2-b^2$ $a^2=17b^2$

4. Find eccentricity

   $e=\frac{c}{a}=\frac{4b}{\sqrt{17}\,b}=\frac{4}{\sqrt{17}}$

5. Match with the options

   $e=\frac{4}{\sqrt{17}}$

   This corresponds to Option C.

Final Answer: $\boxed{\frac{4}{\sqrt{17}}}$