$\Sigma M_{R2} = 0$
$12R_1 + 3(6)(120) = 9(6)(120)$
$R_1 = 360 \, \text{ lb}$
$\Sigma M_{R1} = 0$
$12R_2 = 3(6)(120) + 15(6)(120)$
$R_2 = 1080 \, \text{ lb}$
By ratio and proportion:
$\dfrac{a}{6} = \dfrac{4320}{12}$
$a = 2160 \, \text{ lb}\cdot\text{ft}$
By squared property of parabola:
$\dfrac{b}{6^2} = \dfrac{-8640}{12^2}$
$b = -2160 \, \text{ lb}\cdot\text{ft}$
$EI \, t_{C/A} = (Area_{AC}) \, \bar{X}_C$
$EI \, t_{C/A} = \frac{1}{3}(6)(2160)(\frac{3}{2}) + \frac{1}{2}(12)(4320)(4) - \frac{1}{3}(12)(8640)(3)$
$EI \, t_{C/A} = 6480 \, \text{ lb}\cdot\text{ft}^3$
$EI \, t_{B/A} = (Area_{AB}) \, \bar{X}_B$
$EI \, t_{B/A} = \frac{1}{2}(6a)(2) - \frac{1}{3}(6b)(\frac{3}{2})$
$EI \, t_{B/A} = 6a - 3b$
$EI \, t_{B/A} = 6(2160) - 3(2160)$
$EI \, t_{B/A} = 6480 \, \text{ lb}\cdot\text{ft}^3$
With the values of EI t_{C/A} and EI t_{B/A}, it is obvious that the elastic curve is above point B. The deflection at B (up or down) can also be determined by comparing the values of t_{B/A} and y_{B2}.
By ratio and proportion:
$\dfrac{y_{B2}}{6} = \dfrac{t_{C/A}}{12}$
$y_{B2} = \frac{1}{2}t_{C/A}$
$EI \, y_{B2} = \frac{1}{2}EI \, t_{C/A}$
$EI \, y_{B2} = \frac{1}{2}(6480)$
$EI \, y_{B2} = 3240 \, \text{ lb}\cdot\text{ft}^3$
Since t_{B/A} is greater than y_{B2}, the elastic curve is above point B as concluded previously.
Therefore,
$EI \, \delta_B = EI \, t_{B/A} - EI \, y_{B2}$
$EI \, \delta_B = 6480 - 3240$
$EI \, \delta_B = 3240 \, \text{ lb}\cdot\text{ft}^3$ answer
You can also find the value EI δ_{B} by finding t_{A/C}, t_{B/C}, and y_{B1}. I encourage you to do it yourself.