From the FBD of the whole system

$\Sigma M_E = 0$

$4R_D = 1.5(24)$

$R_D = 9 ~ \text{kN}$

From the FBD of Member AB

$\Sigma M_A = 0$

$3B_V = 2(24)$

$B_V = 16 ~ \text{kN}$

From the FBD of Member BD

$\Sigma F_V = 0$
$C_V + 9 = 16$

$C_V = 7 ~ \text{kN upward}$ *answer*

$\Sigma M_B = 0$

$1.5C_H = 1.5C_V + 3.5R_D$

$1.5C_H = 1.5(7) + 3.5(9)$

$C_H = 28 ~ \text{kN to the right}$ *answer*