If the foci of the ellipse $\large\frac{x^2}{16}+\frac{y^6}{b^2}$$=1$ and the hyperbola. $\large\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide, then the value of $b^2$ is

Question

$\begin{array}{1 1}(A)\;1 \\(B)\;5 \\(C)\;7 \\(D)\;9 \end{array}$