Step-by-step Solution:

\[ \text{ Eccentricity for the ellipse } \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text{ is given by } b^{2}=a^{2}(1-e^{2}) \] \[ \text{For the hyperbola } \frac{x^{2}}{\frac{144}{25}}-\frac{y^{2}}{\frac{81}{25}}=1, \text{ its eccentricity is given by:} \] \[ e_{1} = \sqrt{1 + \frac{b_1^2}{a_1^2}} \] \[ e_{1} = \sqrt{1 + \frac{81}{144}} = \frac{15}{12} \] \[ \text{Foci of the hyperbola: } a_{1} e_{1} = \frac{12}{5} \times \frac{15}{12} = 3 \] \[ \therefore \text{ The focus of the hyperbola is } (3,0) = (ae, 0) \] \[ \text{Since the focus of the ellipse is also at } (ae, 0), \text{ we equate: } 4e = 3 \] \[ \therefore e = \frac{3}{4} \] \[ \text{From the equation } e^{2} = 1 - \left(\frac{b}{a}\right)^{2} \] \[ 1 - e^2 = \frac{b^{2}}{16} \] \[ 1 - \frac{9}{16} = \frac{b^{2}}{16} \] \[ b^{2} = 7 \] Thus, the required value is: \[ \mathbf{b^2 = 7} \]