Solution-
The two parabolas are,
[tex]y=16x^2-c^2 \ and \ y=c^2-16x^2[/tex]
By solving the above two equations we calculate where the two parabolas meet,
[tex]So \ 16x^2-c^2 = c^2-16x^2 \Rightarrow 32x^2=2c^2 \Rightarrow x=\frac{1}{4}c[/tex]
Given the symmetry, the area bounded by the two parabolas is twice the area bounded by either parabola with the x-axis.
[tex]\therefore Area=2\int_{-c}^{c}y.dx= 2\int_{-c}^{c}(16x^2-c^2).dx\\=2[\frac{16}{3}x^3-c^2x]_{-c}^{ \ c}=2[(\frac{16}{3}c^3-c^3)-(-\frac{16}{3}c^3+c^3)]=2[\frac{32}{3}c^3-2c^3]=2(\frac{26c^3}{3})\\=\frac{52c^3}{3}[/tex]
[tex]So \frac{52c^3}{3}=\frac{250}{3}\Rightarrow c=\sqrt[3]{\frac{250}{52}}=1.68[/tex]
