If 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