First, note that these two parabolas are symmetric about the x-axis. The first one has its branches up in y-direction and the second one has its branches down in y-direction. This means that if you want to find area of the region bounded by the parabolas, you can find area of the region bounded by second parabola and x-axis and multiply it by 2.
1. Find x-intercepts of parabola [tex]y=c^2-16x^2.[/tex]
When y=0, you have
[tex]0=c^2-16x^2,\\ \\16x^2=c^2,\\ \\x_1=-\dfrac{c}{4},\ x_2=\dfrac{c}{4}.[/tex]
2. Find area:
[tex]A_1= \int \limits_{-\frac{c}{4}}^{\frac{c}{4}}(c^2-16x^2)\, dx=\left(c^2x-16\dfrac{x^3}{3}\right)\big |^{\frac{c}{4}}_{-\frac{c}{4}}=\\ \\=c^2\left(\dfrac{c}{4}-\left(-\dfrac{c}{4}\right)\right)-\dfrac{16}{3}\left(\dfrac{c^3}{64}-\left(-\dfrac{c^3}{64}\right) \right)=\dfrac{c^3}{2}-\dfrac{c^3}{6}=\dfrac{c^3}{3}.[/tex]
Then the area of the region bounded by the parabolas is
[tex]A=2A_1=2\cdot \dfrac{c^3}{3}.[/tex]
3. Since this area is [tex]\dfrac{250}{3},[/tex] you have
[tex]2\cdot \dfrac{c^3}{3}=\dfrac{250}{3},\\ \\c^3=125,\\ \\c=5.[/tex]
Answer: c=5.
