The increasing order of acidic strength for the aqua complexes $[\text{V(H}_2\text{O)}_6]^{2+}$, $[\text{Fe(H}_2\text{O)}_6]^{2+}$, $[\text{Fe(H}_2\text{O)}_6]^{3+}$, and $[\text{Co(H}_2\text{O)}_6]^{3+}$ is:

The acidic strength of an aqua complex is determined by the polarizing power of the central metal ion, which is directly proportional to its charge and inversely proportional to its ionic radius (charge-to-size ratio).

Higher oxidation states lead to a greater charge density. Therefore, the $+3$ metal ions ($\text{Fe}^{3+}$, $\text{Co}^{3+}$) have a significantly higher polarizing power than the $+2$ metal ions ($\text{V}^{2+}$, $\text{Fe}^{2+}$). This makes the $+3$ aqua complexes more acidic than the $+2$ aqua complexes.

For the $+2$ ions, as we move from left to right across the 3d transition series, the effective nuclear charge increases and the ionic radius generally decreases. The ionic radius of $\text{V}^{2+}$ is larger than that of $\text{Fe}^{2+}$. Consequently, $\text{Fe}^{2+}$ has a higher polarizing power, making $[\text{Fe(H}_2\text{O)}_6]^{2+}$ more acidic than $[\text{V(H}_2\text{O)}_6]^{2+}$.

Similarly, for the $+3$ ions, the ionic radius of $\text{Fe}^{3+}$ is larger than that of $\text{Co}^{3+}$. The smaller size of $\text{Co}^{3+}$ gives it a higher polarizing power, making $[\text{Co(H}_2\text{O)}_6]^{3+}$ more acidic than $[\text{Fe(H}_2\text{O)}_6]^{3+}$.

Combining these factors, the increasing order of acidic strength is:

$[\text{V(H}_2\text{O)}_6]^{2+} < [\text{Fe(H}_2\text{O)}_6]^{2+} < [\text{Fe(H}_2\text{O)}_6]^{3+} < [\text{Co(H}_2\text{O)}_6]^{3+}$

Answer: $[\text{V(H}_2\text{O)}_6]^{2+} < [\text{Fe(H}_2\text{O)}_6]^{2+} < [\text{Fe(H}_2\text{O)}_6]^{3+} < [\text{Co(H}_2\text{O)}_6]^{3+}$