$\Sigma M_{R2} = 0$
$LR_1 = w_o a(\frac{1}{2}a)$
$R_1 = \dfrac{w_oa^2}{2L}$
$M_A = M_c = 0$
$M_B = R_1(L - a) = \dfrac{w_oa^2}{2L}(L - a)$
$L_1 = L - a$
$L_2 = a$
$M_A L_1 + 2M_B(L_1 + L_2) + M_CL_2 + \dfrac{6A_1\bar{a}_1}{L_1} + \dfrac{6A_2\bar{b}_2}{L_2} = 6EI \left( \dfrac{h_A}{L_1} + \dfrac{h_C}{L_2} \right)$
$0 + 2 \cdot \dfrac{w_o a^2}{2L}(L - a) [ \, (L - a) + a \, ] + 0 + 0 + \frac{1}{4}w_o a^3 = 6EI\left( \dfrac{h}{L - a} + \dfrac{h}{a} \right)$
$\dfrac{w_o a^2}{L}(L - a)L + \frac{1}{4}w_o a^3 = 6EI\left[ \dfrac{a + (L - a)}{a(L - a)} \right] \, h$
$w_o a^2(L - a) + \frac{1}{4}w_o a^3 = 6EI\left[ \dfrac{L}{a(L - a)} \right] \, h$
$w_o a^2 L - w_o a^3 + \frac{1}{4}w_o a^3 = \dfrac{6L}{a(L - a)} EI \, h$
$w_o a^2 L - \frac{3}{4}w_o a^3 = \dfrac{6L}{a(L - a)} EI \, h$
$\frac{1}{4}w_o a^2(4L - 3a) = \dfrac{6L}{a(L - a)} EI \, h$
$\dfrac{w_o a^2(4L - 3a)}{4} = \dfrac{6L}{a(L - a)} EI \, h$
$\dfrac{w_o a^3(4L - 3a)(L - a)}{24L} = EI \, h$
Thus,
$EI \, \delta_B = \dfrac{w_o a^3}{24L}(4L - 3a)(L - a)$ answer
