Caroline and Bronya from Tattingstone School sent wonderfully explained solutions to this problem. Caroline drew some diagrams to help her:
Side AB = 1 Perimeter = 3 x 1 = 3 

Side ADB = 4 x ^{1}/_{3} =
^{4}/_{3} Perimeter = 3 x ^{4}/_{3} = 4 

Side ADB = 4 x ^{4}/_{9} =
^{16}/_{9} Perimeter = 3 x ^{16}/_{9} = ^{48}/_{9} = ^{16}/_{3} = 5 ^{1}/_{3} 
She explains:
The pattern is every time you add on a new smaller spike you multiply by 4 and divide by 3.
Here, Bronya describes how she approached the problem:
First of all I looked at the perimeter of the equilateral triangle. It was 3 units. Then I looked at the perimeter of the star. If one side is 1^{1}/_{3} units then the perimeter is 4 units. Then the perimeter of the third shape. I looked at a section like this:
Each section = ^{1}/_{3} + ^{1}/_{9} unit = ^{3}/_{9} + ^{1}/_{9} = ^{4}/_{9} unit
There are 12 sections so the total perimeter = ^{4}/_{9} x ^{12}/_{1} = ^{48}/_{9}
I looked at all the perimeters as ninths:
Perimeter 1 was ^{27}/_{9}
Perimeter 2 was ^{36}/_{9}
Perimeter 3 was ^{48}/_{9}
Each time the perimeter increases by one third.
I think this comes about because in each section a third of the section is added on.The perimeter of the next shape would be ^{64}/_{9} because:
48 divided by 3 is 16 therefore increase ^{48}/_{9} by ^{16}/_{9}
Total would be: ^{48}/_{9} + ^{16}/_{9} = ^{64}/_{9}
Thank you to you both  these are very well reasoned solutions. I wonder whether you could generalise to any shape in this series?