An ellipse has $$OB$$ as semi minor axis, $$F$$ and $$F'$$ its focii and the angle $$FBF'$$ is a right angle. Then the eccentricity of the ellipse is

Take the ellipse in its standard (principal-axis) form with centre at the origin and the major axis along the $$x$$-axis:

$$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\qquad a\gt b\gt 0.$$

The foci are then $$F(c,0)$$ and $$F'(-c,0)$$ where $$c^{2}=a^{2}-b^{2}.$$ The end points of the semi-minor axis are $$B(0,b)$$ and $$B'(0,-b).$$ (We need only the point $$B$$.)

The given condition is that the angle $$FBF'$$ is a right angle: $$\angle FBF' = 90^{\circ}.$$

Vectors along the two sides of this angle are
$$\overrightarrow{BF}= (c-0,\;0-b) = (c,\,-b),$$
$$\overrightarrow{BF'}=(-c-0,\;0-b)=(-c,\,-b).$$

For the angle between these vectors to be $$90^{\circ}$$, their dot product must be zero:

$$\overrightarrow{BF}\cdot\overrightarrow{BF'} = c(-c)+(-b)(-b)= -c^{2}+b^{2}=0.$$

Hence

$$b^{2}=c^{2}\qquad -(1)$$

But for every ellipse $$c^{2}=a^{2}-b^{2}$$. Substituting $$c^{2}=b^{2}$$ from (1):

$$b^{2}=a^{2}-b^{2}\;\;\Longrightarrow\;\;2b^{2}=a^{2}\;\;\Longrightarrow\;\;\frac{b^{2}}{a^{2}}=\frac12.$$

The eccentricity is $$e=\dfrac{c}{a}.$$ Using $$c^{2}=b^{2}$$ and $$b^{2}=a^{2}/2$$ just obtained,

$$c^{2}=a^{2}-b^{2}=a^{2}-\frac{a^{2}}{2}= \frac{a^{2}}{2} \;\;\Longrightarrow\;\; \left(\frac{c}{a}\right)^{2}= \frac12 \;\;\Longrightarrow\;\; e=\frac{1}{\sqrt{2}}.$$

Therefore the eccentricity of the ellipse is $$\dfrac{1}{\sqrt{2}}.$$
Option A which is: $$\frac{1}{\sqrt{2}}$$