Since it is a right pyramid, the vertex V is exactly flush with the centre of the rectangular base. This also means that opposite triangular faces are identical/congruent.
Decorative paper just needs to cover the entire outer surface of the pyramid exactly if we want the minimum.
Entire outer surface area = area of all the triangular faces + area of rectangular base
Draw a line from the midpoint of one breadth of the rectangular base to the centre of the base. Call this midpoint X and the centre C.
Use Pythagoras' Theorem again since triangle VCX is right angled.
CX is parallel to AB but is only half of AB's length. So CX = 15cm.
VC² + CX² = VX²
12² + 15² = VX²
369 = VX²
VX = √369 = √(9x41) = √9√41 = 3√41
Entire outer surface area = rectangular base area + area of 2 triangular faces (length side) + area of 2 triangular faces (breadth side)
= 30cm x 18cm + 2 x ½ x 30cm x 15cm + 2 x ½ x 18cm x 3√41 cm
= 540cm² + 450cm² + 27√41 cm²
= (990 + 27√41) cm²
(Or ≈ 1162.88 cm² = 1160cm² (3s.f) )