Step 1: Understand why these complexes are acidic.
In a metal aqua complex, the central metal ion pulls electron density away from the bound water. This makes it easier for that water to let go of a proton ($\text{H}^+$). The stronger the pull, the more acidic the complex. So we just need to rank how strongly each metal pulls.
Step 2: Find the pulling power.
The pull depends on charge density, which is the charge of the ion divided by its size. A higher charge and a smaller size both raise the charge density and therefore raise the acidity. So we compare charges first, then sizes.
Step 3: Compare by charge.
The two $+2$ ions, $[\text{V}(\text{H}_2\text{O})_6]^{2+}$ and $[\text{Fe}(\text{H}_2\text{O})_6]^{2+}$, have lower charge density and so are the weaker acids. The two $+3$ ions, $[\text{Fe}(\text{H}_2\text{O})_6]^{3+}$ and $[\text{Co}(\text{H}_2\text{O})_6]^{3+}$, have higher charge density and so are the stronger acids.
Step 4: Rank the two $+2$ ions.
Between $\text{V}^{2+}$ and $\text{Fe}^{2+}$, iron sits further right in the period and is a bit smaller with higher effective nuclear charge, so $\text{Fe}^{2+}$ pulls slightly harder than $\text{V}^{2+}$. So $\text{V}^{2+}$ is the least acidic of all.
Step 5: Rank the two $+3$ ions.
Between $\text{Fe}^{3+}$ and $\text{Co}^{3+}$, cobalt is smaller and has a higher charge density, so $\text{Co}^{3+}$ is the most acidic. That puts $\text{Fe}^{3+}$ just below it.
Step 6: Write the full increasing order.
Putting the pieces together from weakest to strongest gives V(II), then Fe(II), then Fe(III), then Co(III).
\[ \boxed{[\text{V}(\text{H}_2\text{O})_6]^{2+} < [\text{Fe}(\text{H}_2\text{O})_6]^{2+} < [\text{Fe}(\text{H}_2\text{O})_6]^{3+} < [\text{Co}(\text{H}_2\text{O})_6]^{3+}} \]