The midspan deflection from
Case No. 8 and
Case No. 11 are respectively,
$\delta = \dfrac{5w_oL^4}{384EI}$ 

$\delta = \dfrac{ML^2}{16EI}$ 

The given beam is transformed into a simple beam with end moment at the right support due to the load at the overhang as shown in the figure below.
EIδ = ½ of EIδ due to uniform load over the entire span  EIδ due to end moment
$EI \, \delta_{midspan} = \dfrac{1}{2}\left[ \dfrac{5(120)(12^4}{384} \right]  \dfrac{2160(12^2)}{16}$
$EI \, \delta_{midspan} = 16\,200  19\,440$
$EI \, \delta_{midspan} = 3\,240$
$EI \, \delta_{midspan} = 3\,240 \, \text{ lb}\cdot\text{ft}^3 \, \text{ upward}$ answer