The correct order of basicity of oxides of vanadium is

${V}_{2}{O}_{3}$:~In,& ${V}_{2}{O}_{3}$, the oxidation state of~$\text{V}$~is as follows

$2 ( x ) + 3 ( - 2 ) = 0$

$⇒ 2x - 6 = 0$

$⇒ x = \dfrac{6}{2} = 3$

The oxidation state of~$V$~in& ${V}_{2}{O}_{3}$~is +3. The vanadium is present in its lowest oxidation state. The vanadium tends to donate the pair of electrons. Hence, it is basic oxide.

${V}_{2}{O}_{5}$:~In& ${V}_{2}{O}_{5}$, the oxidation state of~$V$~is as follows

$2 ( x ) + 5 ( - 2 ) = 0$

$⇒ 2 ( x ) - 10 = 0$

$⇒ x = \dfrac{10}{2} = 5$

The oxidation state of~$V$~in& ${V}_{2}{O}_{5}$~is +5.& The Vanadium tends to donate and accept the pair of electrons.& It is amphoteric oxide.

${V}_{2}{O}_{4}$:~In& ${V}_{2}{O}_{4}$, the oxidation state of~$V$~is as follows

$2 ( x ) + 4 ( - 2 ) = 0$

$⇒ 2 ( x ) - 8 = 0$

$⇒ x = \dfrac{8}{2} = 4$

The oxidation state of~$V$~in& ${V}_{2}{O}_{4}$~is +4.& The Vanadium is present in its other low oxidation state.& The Vanadium tends to donate and accept the pair of electrons.& It is amphoteric oxide and less basic than& ${V}_{2}{O}_{3}$, more basic than~${V}_{2}{O}_{5}$.

So, the correct order of basicity of oxides of vanadium is~${V}_{2}{O}_{3} > {V}_{2}{O}_{4} > {V}_{2}{O}_{5}$.