$\theta = \dfrac{w_oL^3}{48EI}$
$L \theta = \dfrac{w_oL^4}{48EI}$
$t_{A/B} = L \theta$
$t_{A/B} = \dfrac{w_oL^4}{48EI}$
$EI ~ t_{A/B} = \frac{1}{48}w_oL^4$
$\frac{1}{2}(L)(RL)(\frac{2}{3}L) - \frac{1}{3}(L)(\frac{1}{2}w_oL^2)(\frac{3}{4}L) = \frac{1}{48}w_oL^4$
$\frac{1}{3}RL^3 - \frac{1}{8}w_oL^4 = \frac{1}{48}w_oL^4$
$\frac{1}{3}RL^3 = \frac{7}{48}w_oL^4$
$R = \frac{7}{16}w_oL$ answer