The sum of an infinite G.P is 2. If the sum of their squares is $\frac{4}{3}$  then the third term is

$a^2+a^2{r}^2+a^2{r}^4+\cdots \infty =\frac{a^2}{1-r^2}=\frac{4}{3}\Rightarrow 3a^2=4\left(1-r^2\right)$

$\Rightarrow 3a^2=4\left[1-{\left(\frac{2-a}{2}\right)}^2\right]\Rightarrow a=1$

From $\left(1\right),r=\frac{1}{2}$

$\therefore t_3=a{r}^2=1{\left(\frac{1}{2}\right)}^2=\frac{1}{4}$