The foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide. Then the value of $b^2$ is:

For the hyperbola: $\frac{x^2}{144/25}-\frac{y^2}{81/25}=1$
Here $A^2=\frac{144}{25}$ and $B^2=\frac{81}{25}$.
Eccentricity $e_h=\sqrt{1+\frac{B^2}{A^2}}=\sqrt{1+\frac{81}{144}}=\sqrt{\frac{225}{144}}=\frac{15}{12}=\frac{5}{4}$.
Foci of hyperbola: $(\pm A{e}_h,0)=\left(\pm \frac{12}{5}\cdot \frac{5}{4},0\right)=(\pm 3,0)$.
For the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$:$a^2=16$. Foci must be $(\pm 3,0)$, so $a{e}_e=3$.$4e_e=3⟹e_e=\frac{3}{4}$.
We know $b^2=a^2(1-e_e^2)=16\left(1-\frac{9}{16}\right)=16\cdot \frac{7}{16}=7$.