For the discussion of holomorphic tubular neighborhoods and some criteria for their existence see [this question][1].

Let $X$ be a smooth quasi-projective variety over $\mathbb{C}$. Hironaka tells us that there exists a smooth projective compactification $X \subset \bar{X}$, such that $D =\bar{X}\backslash X$ is a strictly normal crossing divisor. Let $D=\cup_{i}^N D_i$ be its decomposition into irreducible smooth divisors. What is the obstruction to finding such $\bar{X}$, such that each $D_i$ has a holomorphic tubular neighborhood in $\bar{X}$?

**Example 1:** Let $S$ be a smooth projective variety, $E$ a vector bundle on $S$. The space $X=\text{Tot}(E)$ is compactified to $\bar{X}=\mathbb{P}_S(E\oplus \mathcal{O}_S)$ with the divisor at infinity $D =\mathbb{P}_S(E)\subset\bar{X}$. Its normal bundle should be $\mathcal{O}_{\mathbb{P}(E)}(1)$ and one should be able to find a holomorphic tubular neighborhood by constructing one in each fiber and gluing together. 

**Example 2:** Using the [answer][2] by [Joey][3], I have convinced myself that if the divisor $D_i$ is $\mathbb{P}^n$ for some $n\geq 2$ and $n+1=\text{dim}(X)$, then it also has such a neighborhood. If $n=1$, then one needs additionally that $(D_i)^2<0$.  

The above examples make me believe that it could be always achievable, as each blow up at a smooth point introduces a new $\mathbb{P}^{n}$ with normal bundle $\mathcal{O}(-1)$.In particular, blowing up seems to only help with the existence of these neighborhoods.


  [1]: https://mathoverflow.net/questions/114414/is-there-any-holomorphic-version-of-the-tubular-neighborhood-theorem
  [2]: https://mathoverflow.net/q/272194
  [3]: https://mathoverflow.net/users/86100/joey