Let $\mu$ be a total extension of Lebesgue measure. For $\lambda>0,$ we define another total extension of Lebesgue measure by $\mu_{\lambda}(X)=\frac{1}{\lambda}(\lambda X).$ We will show there is $\lambda$ such that for all $x \in [0,1],$ there is a Vitali set $V_x$ with $\mu_{\lambda}(V_x)=x.$

In 21:00 - 27:00 of my talk "Paradoxes of perfectly small sets", I construct $X\subset \mathbb{R}$ and fields $F' \subsetneq F$ such that $\mu(\mathbb{R} \setminus X)=0,$ and for any $x, y \in X,$ we have the equivalence $x - y \in F' \Leftrightarrow x-y \in F.$ Fix $\lambda \in F \setminus F'.$ Then $X':=\frac{1}{\lambda} X$ has at most one point per Vitali equivalence class, and $\mu_{\lambda}(X' \cap [0,1))=1.$ Extend $X' \cap [0,1)$ to a Vitali set $V_1.$ Finally, we define

$$V_x = ([0, x) \cap V_1) \cup \left \{y+\frac{1}{2} \bmod 1: y \in [x, 1) \cap V_1 \right \}.$$

Edit: The poster has asked whether the rescaling is necessary. It is. Let $q_n$ enumerate $\mathbb{Q} \cap [0, 1).$ For $x \in (0, 1),$ we define a total extension $\nu$ of Lebesgue measure for which the set of measures of Vitali sets is $(0, x]$, by

$$\nu(A) = \sum_{n=0}^{\infty} x(1- x)^n \mu_{\lambda} (V_1 \cap (A + q_n \bmod 1)).$$

Edit 2: Here's the complete classification of $V(\nu):=\{\nu(Y): Y \text{ a Vitali set}\}$ for total extensions of Lebesgue measure (assuming one exists). For each $\nu,$ there is $x \in (0, 1]$ such that $V(\nu) = [0, x],$ or there is $x \in (0, 1)$ such that $V(\mu) = (0, x].$

My answer above gives constructions for $V(\nu) = [0, 1]$ and $V(\nu) = (0, x].$ Finally, $V(\nu)=[0, x]$ can be achieved by replacing $q_n$ with $q_{n+1}$ in the above construction.

Now we check that these are all the possibilities. First, if $V_1$ is a Vitali set such that $\nu(V_1)=1,$ then our construction of $V_0$ above gives a Vitali set of measure 0. Next, suppose $V_x$ is a Vitali set such that $\nu(V_x) = x,$ and let $\epsilon \in (0, x).$
Let $n$ be such that $\nu(V_x + q_n) < \epsilon$ (addition implicitly taken mod 1). Define a Vitali set
$$U_z=([0, z) \cap V_x) \cup \left \{y+q_n : y \in [z, 1) \cap V_x \right \}.$$
The function $z \mapsto \nu(U_z)$ is Lipschitz of constant 1, so by the intermediate value theorem, there is $z \in (0, 1)$ such that $\nu(U_z) = \epsilon.$

Finally, it remains to verify that there is a Vitali set $V_{\text{max}}$ with $\nu(V_{\text{max}} ) = s:=\sup V(\nu).$ Recursively choose $V_n \in \mathcal{V}_n:=\{\text{Vitali set } V: \nu(V) > (1 - 3^{-n})s\}$ such that
$$\nu(V_{n+1} \triangle V_n) < (1 + 5^{-n})\inf_{U \in \mathcal{V}_{n+1}} \nu(U \triangle V_n).$$

Let $W = \liminf V_n.$ It is clear that $W$ has at most one point per Vitali equivalence class and can thus be extended to a Vitali set $V_{\text{max}}.$ That $\nu(V_{\text{max}})\ge \nu(W) \ge \nu(V_n)$ for all $n$ can be verified by checking
$$\nu(\{x \in V_i: \{x+q_m, x+ q_n\} \subset \limsup V_n \})=0$$
for each $m<n$ and
$$\nu(\{x \in V_i: |\{m: x+q_m \in \bigcup_{n<\omega} V_n\}|=\aleph_0\})=0.$$