Given that the length of the cube's edges is [tex]s[/tex], then the length of the diagonal along any face of the cube is, by the Pythagorean theorem,
[tex]\sqrt{s^2+s^2}=s\sqrt2[/tex]
Then the diagonal along the opposite vertices of the cube (i.e. the line segment through the center of the cube joining its opposite corners) is, again by the Pythagorean theorem,
[tex]\sqrt{(s\sqrt2)^2+s^2}=s\sqrt3[/tex]
The radius of the sphere, [tex]r[/tex], corresponds to half the length of the cube's diagonal; that is,
[tex]r=\dfrac{\sqrt3}2s[/tex]
The surface area of the cube in terms of its edge length [tex]s[/tex] is
[tex]A=6s^2[/tex]
which means the area, rewritten in terms of the sphere's radius [tex]r[/tex], is
[tex]A=6\left(\dfrac2{\sqrt3}r\right)^2=8r^2[/tex]
Hence the rate of change of the cube's surface area with respect to a change in the sphere's radius is
[tex]\dfrac{\mathrm dA}{\mathrm dr}=16r[/tex]
When [tex]r=2[/tex], we have
[tex]\dfrac{\mathrm dA}{\mathrm dr}\bigg|_{r=2}=16(2)=32[/tex]
while when [tex]r=8[/tex], we get
[tex]\dfrac{\mathrm dA}{\mathrm dr}\bigg|_{r=8}=16(8)=128[/tex]
I'm not sure what part (d) reads, so I'll leave that to you...
