Solving for reactions

$\Sigma M_{R2} = 0$

$6R_1 = 80(4)(4)$

$R_1 = 213.33 \, \text{ lb}$

$\Sigma M_{R1} = 0$

$6R_2 = 80(4)(2)$

$R_2 = 106.67 \, \text{ lb}$

From the conjugate beam

$\Sigma M_A = 0$

$6F_2 + \frac{1}{3}(4)(640) [\frac{3}{4}(4)] = \frac{1}{2}(4)(853.33) [\frac{2}{3}(4)] + \frac{1}{2}(2)(213.33) [4 + \frac{1}{3}(2)]$

$F_2 = 497.77 \, \text{ lb}\cdot\text{ft}^2$

$M_B = \frac{1}{2}(2)(213.33 [\frac{1}{3}(2)] - 2F_2$

$M_B = \frac{1}{2}(2)(213.33 [\frac{1}{3}(2)] - 2(497.77)$

$M_B = -853.32 \, \text{ lb}\cdot\text{ft}^3$

Thus, the deflection at B is

$EI \, \delta_B = M_B$

$EI \, \delta_B = -853.32 \, \text{ lb}\cdot\text{ft}^3$

$EI \, \delta_B = 853.32 \, \text{ lb}\cdot\text{ft}^3$ downward *answer*