In general the answer is **no**.

Take $X=\mathbb{A}^3$ with coordinates $x,y,z$ and let $V_1$ and $V_2$ be two lines meeting in one point, for instance

$V_1:=\{x=y=0\}, \quad V_2:= \{x=z=0\}$.

Then the ideal of $V_1 \cup V_2$ is $I=(x, yz)$ and the equation of the blow-up $\widetilde{X}$ of $X$ along $I$ are given in $X \times \mathbb{P}^1$ by

$\lambda x - \mu yz=0$,

where $[\lambda : \mu]$ are homogeneous coordinates in $\mathbb{P}^1$. In the chart $\mu=1$ the blow-up is then given by

$\textrm{Spec }k[x,y,z, \lambda]/(\lambda x - yz)$,

hence it has an isolated singularity at the origin.

The other chart $\lambda=1$ is instead smooth, so this is actually the unique singular point of $\widetilde{X}$.