From the FBD of the whole frame
$\dfrac{y}{2}= \dfrac{1.5}{1}$
$y = 3 ~ \text{m}$
$\Sigma F_x = 0$
$A_H = 0$
$\Sigma M_A = 0$
$3R_F = 6(240)$
$R_F = 480 ~ \text{kN}$
$\Sigma M_F = 0$
$3A_V = 3(240)$
$A_V = 240 ~ \text{kN}$
From the FBD of member BG
$\Sigma M_B = 0$
$2E_V = 4(240)$
$E_V = 480 ~ \text{kN}$
$\Sigma M_E = 0$
$2B_V = 2(240)$
$B_V = 240 ~ \text{kN}$
From the FBD of member CF
$\Sigma F_y = 0$
$C_V + 480 = 480$
$C_V = 0$
$\Sigma M_E = 0$
$3C_H = 1(240)$
$C_H = 80 ~ \text{kN}$
$\Sigma M_C = 0$
$3E_H = (3 - 1)(240)$
$E_H = 160 ~ \text{kN}$
From the FBD of member AC
$\Sigma M_A = 0$
$1.5B_H + 1(240) = 4.5(80)$
$B_H = 80 ~ \text{kN}$
Summary
$B_H = 80 ~ \text{kN}$ and $B_V = 240 ~ \text{kN}$
$C_H = 80 ~ \text{kN}$ and $C_V = 0$
$E_H = 160 ~ \text{kN}$ and $E_V = 480 ~ \text{kN}$
Thus,
$R_B = \sqrt{80^2 + 240^2} = 252.98 ~ \text{kN}$ ← answer
$R_C = \sqrt{80^2 + 0^2} = 80 ~ \text{kN}$ ← answer
$R_E = \sqrt{160^2 + 480^2} = 505.96 ~ \text{kN}$ ← answer
