If the points of intersection of the ellipse x ^ 2 16 + y ^ 2 b ^ 2...

${y}^{2} = 3{x}^{2}$ and ${x}^{2} + {y}^{2} = 4b$

Solving both the equations we get,

${x}^{2} = b$

As $\dfrac{{x}^{2}}{16} + \dfrac{{y}^{2}}{{b}^{2}} = 1$

$⇒ \dfrac{{x}^{2}}{16} + \dfrac{3{x}^{2}}{{b}^{2}} = 1$

$⇒ \dfrac{b}{16} + \dfrac{3}{b} = 1$

$⇒ {b}^{2} - 16b + 48 = 0$

$⇒ \left(b - 12\right)\left(b - 4\right) = 0$

$⇒ b = 12\left\{∵ b > 4\right\}$