If the CFSE of [Ti ( H _ 2 O) _ 6 ] ^ 3 + is - 96 . 0kJ / mol ,...

In ${\left[Ti{\left({H}_{2}O\right)}_{6}\right]}^{3 +}$, oxidation state of Titanium is +3.
Thus, electronic configuration:

${Ti}^{+ 3}: 3{d}^{1} ⇒ {t}_{2g}^{1, 0, 0}, {{e}_{g}}^{0, 0}$

We know CFSE $= \left[- 0 . 4{n}_{{t}_{2g}} + 0 . 6{n}_{{e}_{g}}\right]\Delta o$ ( $\Delta o = Octahedralsplittingenergy$)

$\Delta o = \dfrac{96 \times {10}^{3}}{0 . 4 \times 6 \times {10}^{23}} = 4 \times {10}^{- 19}J$

Using planck's equation and fact that octahedral splitting energy per complex is:
$⇒ \Delta o = \dfrac{hc}{\lambda} = 4 \times {10}^{- 19}J$

$⇒ \lambda = \dfrac{6 . 4 \times {10}^{- 34} \times 3 \times {10}^{8}}{4 \times {10}^{- 19}} = 4 . 8 \times {10}^{- 7}m$

$⇒ \lambda = 480 \times {10}^{- 9}m$

$∴ \lambda = 480nm$