If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse is :

For an ellipse centred at the origin with the major axis along the $$x$$-axis, we use the standard form

$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\qquad a \gt b\,(\,\text{major}\, \gt \text{minor}\,)$$

Eccentricity: $$e=\sqrt{1-\frac{b^{2}}{a^{2}}}$$
Distance between the two foci: $$2ae$$
Length of the minor axis: $$2b$$

The question states: “length of the minor axis is one fourth of the distance between the foci”.
Translate this directly into an equation:

$$2b=\frac{1}{4}\times(2ae)$$

Simplify:

$$2b=\frac{ae}{2}\qquad\Longrightarrow\qquad4b=ae$$ $$-(1)$$

Replace $$b$$ by its expression in terms of $$a$$ and $$e$$. From the definition of eccentricity,

$$b=a\sqrt{1-e^{2}}$$ $$-(2)$$

Substitute $$(2)$$ into $$(1)$$:

$$4\bigl(a\sqrt{1-e^{2}}\bigr)=ae$$

Assuming $$a\neq0$$, divide both sides by $$a$$:

$$4\sqrt{1-e^{2}}=e$$

Square both sides to eliminate the square root:

$$16(1-e^{2})=e^{2}$$

Expand and collect like terms:

$$16-16e^{2}=e^{2}\qquad\Longrightarrow\qquad17e^{2}=16$$

Solve for $$e$$:

$$e^{2}=\frac{16}{17}\qquad\Longrightarrow\qquad e=\frac{4}{\sqrt{17}}$$

Since $$e$$ must be positive and less than 1, this value is acceptable.
Hence, the eccentricity of the ellipse is $$\frac{4}{\sqrt{17}}$$.

Option A is correct.