Given:
The original surface area of the cuboid is 200 square units.
The formula for the surface area(A) of a cuboid is,
[tex]A=2(lh+lw+hw)[/tex]We were told the original surface area is being dilated by 3,
Therefore, all the sides of the cuboid will be increased by 3.
[tex]\begin{gathered} A_2=2(3l\times3h+3l\times3w+3h\times3w) \\ A_2=2(9lh\times9lw+9wh) \end{gathered}[/tex]Let us now factorize out 9 from the formula above
[tex]A_2=9(2(lh+lw+wh))[/tex]Recall that
[tex]A=2(lh+lw+hw)[/tex]Therefore,
[tex]A_2=9A[/tex]Also, recall that the original surface area of the cuboid is 200 square units.
[tex]A=200\text{unit}^2[/tex]Hence,
[tex]A_2=9\times200unit^2=1800unit^2[/tex]Therefore, the surface area of the dilated image is
[tex]1800\text{unit}^2[/tex]