Given, ${\left(\sum _{i=1}^3\frac{a_i}{2^i}\right)}^2=\sum _{i=1}^3\frac{a_i^2}{2^i}$

$\Rightarrow {\left(\frac{a_1}{2}+\frac{a_2}{4}+\frac{a_3}{8}\right)}^2=\frac{a_1^2}{2}+\frac{a_2^2}{4}+\frac{a_3^2}{8}$

$\Rightarrow \frac{a_1^2}{4}+\frac{a_2^2}{16}+\frac{a_3^2}{64}+\frac{a_1{a}_2}{4}+\frac{a_2{a}_3}{16}+\frac{a_3{a}_1}{8}$

$=\frac{a_1^2}{2}+\frac{a_2^2}{4}+\frac{a_3^2}{8}$

$\Rightarrow \frac{a_1^2}{4}+\frac{3a_2^2}{16}+\frac{7a_3^2}{64}=\frac{a_1{a}_2}{4}+\frac{a_2{a}_3}{16}+\frac{a_3{a}_1}{8}$

$\Rightarrow 16a_1^2+12a_2^2+7a_3^2$

$=16a_1{a}_2+4a_2{a}_3+8a_3{a}_1$

$\Rightarrow \left(8a_1^2+8a_2^2-16a_1{a}_2\right)+\left(4a_2^2+a_3^2-4a_2{a}_3\right)+\left(2a_3^2+8a_1^2-8a_3{a}_1\right)+4a_3^2=0$

$\Rightarrow 8{\left(a_1-a_2\right)}^2+{\left(2a_2-a_3\right)}^2+2{\left(a_3-2a_1\right)}^2+4a_3^2=0$

The above relation will be true if 

$a_1=a_2,2a_2=a_3,a_3=2a$ and $a_3=0$

So, $a_1=a_2=a_3=0$

Then, set $A$ contains exactly one element.