We are given that the length of the minor axis is equal to one fourth of the distance between the foci.
Let the length of the minor axis be \( b \), and the length of the major axis be \( a \).
The distance between the foci is \( 2c \), where \( c \) is the distance from the center to the foci.
Given:
\[
b = \frac{1}{4} \times 2c = \frac{c}{2}
\]
Now, the relationship between \( a \), \( b \), and \( c \) in an ellipse is:
\[
c^2 = a^2 - b^2
\]
Substitute \( b = \frac{c}{2} \) into the equation:
\[
c^2 = a^2 - \left(\frac{c}{2}\right)^2
\]
\[
c^2 = a^2 - \frac{c^2}{4}
\]
Multiplying both sides by 4 to eliminate the fraction:
\[
4c^2 = 4a^2 - c^2
\]
\[
5c^2 = 4a^2
\]
\[
c^2 = \frac{4a^2}{5}
\]
Now, the eccentricity \( e \) of an ellipse is defined as:
\[
e = \sqrt{1 - \frac{b^2}{a^2}}
\]
Substitute the value of \( b \) and \( c \) into the equation:
\[
e = \sqrt{1 - \frac{\left(\frac{c}{2}\right)^2}{a^2}} = \sqrt{1 - \frac{c^2}{4a^2}}
\]
Substitute \( c^2 = \frac{4a^2}{5} \) into the equation:
\[
e = \sqrt{1 - \frac{\frac{4a^2}{5}}{4a^2}} = \sqrt{1 - \frac{1}{5}}
\]
\[
e = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}
\]
Thus, the eccentricity is \( e = \frac{4}{\sqrt{17}} \).