Hello.
Using the Pythagora's Theorem we can find CE. $$CE=\sqrt{2^2-1^2}=\sqrt{3}.$$
Now, AE=AC+CE, this implies that: $$AC=AE-CE=6\sqrt{3}-\sqrt{3}=5\sqrt{3}.$$
To calculate AB, we will use that triangles ABC and EDC are similar. So, $$\frac{AB}{ED}=\frac{AC}{CE},\ this\ implies\ that,\ \frac{AB}{1}=\frac{5\sqrt{3}}{\sqrt{3}},\ that\ is,\ AB=5.$$
Now, we can calculate BC using the Pythagora's Theorem again. We will get, $$BC=\sqrt{5^2+\ \left(5\sqrt{3}\right)^2}=\sqrt{25+75}=\sqrt{100}=10.$$
In conclusion, the perimeter of ABC is $$15+5\sqrt{3}.$$
The correct answer is D.
I hope this can help you.
Feel free to ask me again if you have any doubt.
