Given $f(x)=a_0+a_1 x+a_2 x^2+\ldots+a_n x^n, a_0, a_1, a_2$, $\ldots a_n \in Z$. ...(i)
Since $(3+i)$ and $(2-\sqrt{3})$ are roots of equation (i)
Hence $(3-i)$ and $(2+\sqrt{3})$ will also be the roots of given equation.
Therefore, now we have at least 4 roots of given equation. In other words, then the least value of $n$ will be 4 .
$$
\therefore n=4 \text {. }
$$

If 4 roots are there, then the correspondingly equation (i) will reduce to:
$$
\begin{array}{r}
x^4-(\alpha+\beta+\gamma+\delta) x^3+(\alpha \beta+\beta \gamma+\gamma \delta+\delta \alpha+\alpha \gamma+\beta \delta) x^2 \\
+(\alpha \beta \gamma+\beta \gamma \delta+\gamma \delta \alpha+\alpha \beta \delta) x+\alpha \beta \gamma \delta=0 \ldots \text { (ii) }
\end{array}
$$

Clearly, $\alpha \beta \gamma \delta$ is the constant term of equation
Hence $a_0=\alpha \beta \gamma \delta=(3+i)(3-i)(2-\sqrt{3})(2+\sqrt{3})$
$$
\Rightarrow a_0=10
$$