The surface area of a cone of slant length L and base radius R is πRL. This doesn't include the base area of πR^2. So the surface area of the frustum is π(R1L1-R2L2) where R1, R2 are the top and base radii and L1, L2 are the full slant lengths of the cones. It's assumed that L1=7 for the right cone. We know that the ratio R2/R1=3/7. This ratio applies to L2/L1 and the heights of the nested cones H1, H2.
So we can substitute R1=7, R2=3, L1=7, L2=3L1/7=3. The surface area of the frustum without the extra material for the seam is π(49-9)=40π. If we assume the extra material for the seam forms a truncated sector with a base extension of ½", we need to calculate how much material this is.
The sector length at the base of the original cone is S=Rø where ø is the angle of the sector. S=½" and R=R1=7 so ø=1/14. The area of the sector is øR1^2=49/14. For the smaller cone the sector area is øR2^2=9/14 and the difference is the area of extra material we need=40/14=20/7 sq in. Therefore the total amount of material is 40π+20/7 sq in=128.52 sq in.