Answer
$F_D=11.5KN$
Work Step by Step
We know that
$F_x=\frac{dm}{dt}(v_{Bx}-v_{Ax})$
$\implies B_x=50(1.5cos30-0)$
$\implies B_x=64.952N$
Now $\Sigma F_s=m\frac{dv}{dt}-\frac{dm}{dt}v_{D/e}+\frac{dm_i}{dt}v_{D/i}$
$\implies-F_D=0-v_{D/e}(\frac{dm}{dt}+q\rho_{\circ})+v_{D/i} (q\rho_{\circ})$
We plug in the known values to obtain:
$F_D=450(0.4+1.22(50))-263.9(1.22)(50)$
This simplifies to:
$F_D=11.5KN$
