Regular expression given is $(a+b)^*(aa+bb)(a+b)^*$ , means all strings having double letter
DFA
Option B,
$R_2=\\ \left\{\underbrace{\underbrace{a}_{reach \;to\;q_1}\underbrace{(ba)^*}_{still\; at\;q_1}\underbrace{(a+bb)}_{reach\;to\;q_3}+\underbrace{b}_{reach \;to\;q_2}\underbrace{(ab)^*}_{still\; at\;q_2}\underbrace{(b+aa)}_{reach\;to\;q_3}}_{we\; are\; at \;state\; q_3}\right \} \underbrace{(a+b)^*}_{at\; q_3\; with\; anything\;including\;\epsilon}$
is correct as per DFA .
Another way is to do is using state elimination
Remove Edges in between $q_1$ and $q_2$ and then states, we will get the exact option
Note :
1. In option A, it is wrong because of $(a+b)^+$ , once we are at state $q_3$, then we need minimum $\epsilon$ to reach $q_3$ , that is missing in option A
2. In option C, $R_3$ can generate $\epsilon, a $ and $b$, but we need minimum $aa$ or $bb$ to reach final and as per given regular expression.
3. In option D, $R_4$ can generate $ab$ or $ba$, that should not be .