Suppose that Box-I contains 8 red, 3 blue and 5 green balls, Box-ll...

$P\left(\text{Red from Box}I\right) = \dfrac{8}{16} = \dfrac{1}{2}$

$P\left(\text{Blue from Box}I\right) = \dfrac{3}{16}$

$P\left(\text{Green from Box}I\right) = \dfrac{10}{32} = \dfrac{5}{16}$

Now,

$P\left(\text{Green from Box}II\right) = \dfrac{15}{48}$

$P\left(\text{Green from Box}III\right) = \dfrac{12}{16} = \dfrac{3}{4}$

$P\left(\text{White from Box}IV\right) = \dfrac{6}{32} = \dfrac{3}{16}$

Event $A$ : One of the chosen ball is white
$B:$ At least one of the chosen ball is green

$P\left(\dfrac{A}{B}\right) = \dfrac{P\left(A \cap B\right)}{P\left(B\right)}$

$= \dfrac{\dfrac{5}{16} \times \dfrac{3}{16}}{\dfrac{1}{2} \times \dfrac{15}{48} + \dfrac{3}{16} \times \dfrac{3}{4} + \dfrac{5}{16} \times 1}$

$= \dfrac{5}{52}$