What is the quotient $\mathbb Z[\sqrt{3}]/(1+2\sqrt{3})$?
Notice that: $(-1+2\sqrt{3})(1+2\sqrt{3})=11\in I$ Also $(1+2\sqrt{3})\sqrt{3}=6+\sqrt{3}\in I$. Thus $a+b\sqrt{3}+I=a+b(-6)+I=a-6b+I=[(a-6b)\mod\,11]+I$. Hence: $$\mathbb{Z}[\sqrt{3}]/I=\{a+I:0\leq a\leq 10\}$$
Now Consider $\phi:\mathbb{Z}_{11}\rightarrow \mathbb{Z}[\sqrt{3}]/I$ that sends $x$ to $x+I$. Verify that $\phi$ is a surjective group homomorphism. Thus, $|\mathbb{Z}[\sqrt{3}]/I|$ divides $11$ ......
Finally, show that $|\mathbb{Z}[\sqrt{3}]/I|\not=1$ by showing that $1\not\in I$ using a norm argument.
There is a general method for computing these quotients, which is quite straight forward when you feel comfortable with polynomials.
Since $\mathbb{Z}[\sqrt{3}] \cong \mathbb{Z}[x]/(x^2-3)$, we have $Q:=\mathbb{Z}[\sqrt{3}]/(1+2 \sqrt{3}) \cong \mathbb{Z}[x]/(x^2-3,1+2x)$. In $Q$ we have $0=(2x+1)(2x-1)=4 x^2-1=11$. Therefore
$Q \cong \mathbb{F}_{11}[x]/(x^2-3,2 \cdot 6 + 2 \cdot x) = \mathbb{F}_{11}[x]/(x^2-3,6+x)=\mathbb{F}_{11}/(5^2-3)=\mathbb{F}_{11}.$
In questions like this it can help to know the order of the quotient ring a priori. For this, one can use the following argument:
One has the chain of ideals $(11) = (1+2\sqrt{3})(1-2\sqrt{3}) \subset (1+2\sqrt{3}) \subset \mathbb Z[\sqrt{3}].$
Multiplication by $(1+2\sqrt{3})$ induces an isomorphism between $\mathbb Z[\sqrt{3}]/(1-2\sqrt{3})$ and $(1+ 2\sqrt{3})/(11)$.
Galois conjugation (swapping $\sqrt{3}$ and $-\sqrt{3}$) gives an isomorphism betweem $\mathbb Z[\sqrt{3}]/(1-2\sqrt{3})$ and $\mathbb Z[\sqrt{3}]/(1+2\sqrt{3}).$
Putting all this together, we find that the order of $\mathbb Z[\sqrt{3}]/(11)$ is equal to the square of the order of $\mathbb Z[\sqrt{3}]/(1+2\sqrt{3}).$ Since the former is isomorphic (as an abelian group) to $\mathbb Z/(11) \times \mathbb Z/(11)$, it has order $11^2$, and so $\mathbb Z[\sqrt{3}]/(1+2\sqrt{3})$ has order $11$.
In this particular case, we are done, since the quotient ring has order $11$ which is prime, and so is necessarily the prime field of order $11$.
In general, one can show that if $a$ and $b$ are coprime, then $\mathbb Z[\sqrt{3}]/(a + b \sqrt{3})$ is isomorphic to $\mathbb Z/(a^2 - 3b^2)$, and it is a good exercise to write down the details, using the same argument as in Amr's answer (combined with the obvious generalization of the above argument, to see that order of the quotient ring is equal to $|a^2 - 3b^2|$).
If you want to see another phrasing of essentially the same argument (in the context of the ring $\mathbb Z[i]$, but it goes exactly the same way) see this answer.