Your states can’t be made to work. When you’re in the initial state and read $a_1$, you’ve no way to know whether $a_1\ldots a_n>b_1\ldots b_n$, $a_1\ldots a_n=b_1\ldots b_n$, or $a_1\ldots a_n<b_1\ldots b_n$, so you’ve no way to know what the appropriate transition is.
If $a_1\ldots a_n\ne b_1\ldots b_n$, there is a smallest $k\in\{1,\ldots,n\}$ such that $a_k\ne b_k$, and it’s not hard to see that $a_1\ldots a_n>b_1\ldots b_n$ if and only if $a_k>b_k$. Using that idea, I can construct a suitable DFA with five states; $q_0$ is the initial state, $q_4$ is the only acceptor state, and the transitions are as follows:
$$\begin{array}{}
q_0\overset{0}\longrightarrow q_1&&q_0\overset{1}\longrightarrow q_2\\
q_1\overset{0}\longrightarrow q_0&&q_1\overset{1}\longrightarrow q_3\\
q_2\overset{0}\longrightarrow q_4&&q_2\overset{1}\longrightarrow q_0\\
q_3\overset{0}\longrightarrow q_3&&q_3\overset{1}\longrightarrow q_3\\
q_4\overset{0}\longrightarrow q_4&&q_4\overset{1}\longrightarrow q_4
\end{array}$$
After reading $a_1b_1\ldots a_kb_k$ for any $k$, the automaton will be in state $q_0$ if $a_1\ldots a_k=b_1\ldots b_k$. If $k$ is minimal such that $a_k\ne b_k$, then either $a_kb_k=10$ or $a_kb_k=01$. In the first case the automaton is in state $q_4$; in the second it’s in state $q_3$. Both are trap states, so in the first case the input is accepted, and in the second it is not.
I constructed this DFA on the assumption that the length of the input string would be even; if it was supposed to test for strings of odd length and reject them, some modification would be required.
