1. Find the eccentricities $e_1$ and $e_2$

Given ellipse: $\frac{x^2}{b^2}+\frac{y^2}{25}=1, \quad b<5$ Since $25>b^2$, the major axis is along the $y$-axis. Thus, $a^2=25,\quad b^2=b^2$ For the ellipse, $e_1=\sqrt{1-\frac{b^2}{25}}$

Given hyperbola: $\frac{x^2}{16}-\frac{y^2}{b^2}=1$ Here, $a^2=16,\quad b^2=b^2$ For the hyperbola, $e_2=\sqrt{1+\frac{b^2}{16}}$

2. Use the condition $e_1e_2=1$

$\sqrt{1-\frac{b^2}{25}}\cdot \sqrt{1+\frac{b^2}{16}}=1$ Squaring both sides, $\left(1-\frac{b^2}{25}\right)\left(1+\frac{b^2}{16}\right)=1$ Expand: $1+\frac{b^2}{16}-\frac{b^2}{25}-\frac{b^4}{400}=1$ $\frac{b^2}{16}-\frac{b^2}{25}-\frac{b^4}{400}=0$ Take LCM of first two terms: $\frac{25b^2-16b^2}{400}-\frac{b^4}{400}=0$ $\frac{9b^2-b^4}{400}=0$ $b^2(9-b^2)=0$ Since $b>0$ for a valid conic, $b^2=9 \implies b=3$

3. Find the foci of the given ellipse and hyperbola

For the ellipse: $c_1^2=a^2-b^2=25-9=16 \implies c_1=4$ Since major axis is along $y$-axis, its foci are: $(0,\pm 4)$

For the hyperbola: $c_2^2=a^2+b^2=16+9=25 \implies c_2=5$ Since transverse axis is along $x$-axis, its foci are: $(\pm 5,0)$

So the four points are: $(\pm 5,0),\quad (0,\pm 4)$

4. Equation of the required ellipse

The ellipse has axes along the coordinate axes, so its equation is of the form $\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$ It passes through $(\pm 5,0)$ and $(0,\pm 4)$. Hence, $A=5,\quad B=4$ So the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$

5. Find its eccentricity

For this ellipse, $a^2=25,\quad b^2=16$ Therefore, $e=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$

6. Check options

The correct option is: $\boxed{\frac{3}{5}}$ which is Option B.