does anyone know the surface area for this problem. if so i will mark brainliest.

From the given volume, we know that the ratio is:
[tex]\text {Ratio of yellow eraser to white eraser = } \sqrt[3]{ \dfrac{16}{2} } [/tex]

From the given area, we know that the ratio is:
[tex]\text {Ratio of yellow eraser to white eraser = } \sqrt{ \dfrac{52}{x} } [/tex]

Equate the two ratio and solve for x:
[tex]\sqrt[3]{ \dfrac{16}{2} } = \sqrt{ \dfrac{52}{x} } [/tex]

Cube both sides:
[tex]\dfrac{16}{2} = \bigg(\sqrt{ \dfrac{52}{x} }\bigg)^3[/tex]

Square both sides:
[tex]\bigg(\dfrac{16}{2} \bigg)^2 = \bigg( \dfrac{52}{x} }\bigg)^3[/tex]

Simplify each term:
[tex]\dfrac{16^2}{2^2} = \dfrac{52^3}{x^3} [/tex]

Cross multiply:
[tex]256x^3 = 562432[/tex]

Divide both sides by 256:
[tex] x^3 = 2197[/tex]

Cube root both sides:
[tex] x = 13[/tex]