The spin-only magnetic moment value of the most basic oxide of...

${V}_{2}{O}_{3}$ is the most basic oxide among the given oxides.

In ${V}_{2}{O}_{3}$, Vanadium is in $+ 3$ oxidation state

${V}^{3 +} ⇒ \left[Ar\right]3{d}^{2}4{s}^{0}$

So, $2$ unpaired electrons are there,

Its magnetic moment is equal to $\mu = \sqrt{n\left(n + 2\right)}BM = \sqrt{2\left(2 + 2\right)}BM$

$= 2 . 83BM ≃ 3BM$

($n =$ number of unpaired electrons)