1. Set up the ellipse

   Take the standard ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \qquad a>b$ with foci $S(c,0),\quad S'(-c,0),\quad \text{where } c^2=a^2-b^2.$

   An extremity of the minor axis is $B(0,b).$

2. Use the right angle condition

   Given that $\triangle S'BS$ is right-angled at $B$, so $\overrightarrow{BS}\cdot \overrightarrow{BS'}=0.$

   Now, $\overrightarrow{BS}=(c,-b),\qquad \overrightarrow{BS'}=(-c,-b).$

   Their dot product is $(c)(-c)+(-b)(-b)=-c^2+b^2.$

   For perpendicularity, $-c^2+b^2=0 \implies b^2=c^2 \implies b=c.$

3. Use the area condition

   The base $S'S$ has length $2c$, and the perpendicular distance of $B$ from the major axis is $b$.

   So area of $\triangle S'BS$ is $\frac12 \cdot 2c \cdot b = bc.$

   Given area $=8$, $bc=8.$

   Since $b=c$, $b^2=8 \implies b=2\sqrt2.$

   Hence $c=2\sqrt2.$

4. Find $a$

   Using $c^2=a^2-b^2,$ and $c^2=b^2=8$, $8=a^2-8 \implies a^2=16 \implies a=4.$

5. Length of latus rectum

   For the ellipse, length of a latus rectum is $\frac{2b^2}{a}.$

   Therefore, $\frac{2\cdot 8}{4}=4.$

6. Check options

   $4$ corresponds to Option C.

Final Answer: $\boxed{4}$