$\Sigma M_{R2} = 0$
$6R_1 = 80(4)(4)$
$R_1 = \frac{640}{3} \, \text{ lb}$
$\Sigma M_{R1} = 0$
$6R_2 = 80(4)(2)$
$R_2 = \frac{320}{3} \, \text{ lb}$
$t_{A/C} = \dfrac{1}{EI}(Area_{AC}) \, \bar{X}_A$
$t_{A/C} = \dfrac{1}{EI} [ \, \frac{1}{2}(4)(2560/3)(\frac{8}{3}) + \frac{1}{2}(2)(\frac{640}{3})(4 + \frac{2}{3}) - \frac{1}{3}(4)(640)(3) \, ]$
$t_{A/C} = \dfrac{8960}{3EI}$
$t_{B/C} = \dfrac{1}{EI}(Area_{BC}) \, \bar{X}_B$
$t_{B/C} = \dfrac{1}{EI} [ \, \frac{1}{2}(2)(\frac{640}{3})(\frac{2}{3}) \, ]$
$t_{B/C} = \dfrac{1280}{9EI}$
By ratio and proportion:
$\dfrac{y}{2} = \dfrac{t_{A/C}}{6}$
$y = \dfrac{2}{6}\left( \dfrac{8960}{3EI} \right)$
$y = \dfrac{8960}{9EI}$
$\delta_B = y - t_{B/C}$
$\delta_B = \dfrac{8960}{9EI} - \dfrac{1280}{9EI}$
$\delta_B = \dfrac{2560}{3EI}$
$EI \, \delta_B = \frac{2560}{3} \, \text{ lb}\cdot\text{ft}^3$ answer
