Let the four-digit number be $\text{aabb}.$
The number we can write $1000 \; \text{a} + 100 \; \text{a} + 10 \; \text{b} + \text{b}. = 1100 \; \text{a} + 11 \; \text{b} = 11(100 \; \text{a} + \text{b}) \longrightarrow(1)$
This number will be perfect square if a number in the form of $\text{a0b}$ is divisible by $11$ and when $\text{a0b}$ is divided by $11$ perfect square will be get.
$\text{a0b}$ will be divisible by $11$ when $a+b=11$
$\boxed{\text{ b=11-a }}$
put the value of $\text{a}$ in equation $(1).$
$\Rightarrow 11(100 \; \text{a} + 11 – \text{a})$
$\Rightarrow 11(99 \; \text{a} + 11)$
$\Rightarrow 11 \times 11(9 \; \text{a} + 1)$
$\Rightarrow 121 \times(9 \; \text{a} + 1)$
$\Rightarrow 121 \times 64 = 7744.$
Correct Answer $: \text{D}$