Answer
$F_{D} = C_{D} A \rho V^{2} $
Work Step by Step
Let's assume arbitrary exponents:
$F_{D} = C_{D} A^{\alpha} \rho^{\beta} V^{\gamma} $
With LMT (length-mass-time) dimensional analysis:
$F = ma \rightarrow M \times LT^{-2}$
$A \rightarrow L^{2}$
$\rho = \frac{m}{V} \rightarrow M \times L^{-3} $
$V = \frac{ds}{dt} \rightarrow L \times T^{-1}$,
therefore
$MLT^{-2} = L^{2\alpha} \times M^{\beta}L^{-3\beta} \times L^{\gamma}T^{-\gamma} $
Which yields, by exponent comparison:
$M: \beta = 1$
$T: -\gamma = -2 \rightarrow \gamma = 2$
$L: 2\alpha + \gamma - 3\beta = 1 \rightarrow \alpha = 1$
In conclusion:
$F_{D} = C_{D} A \rho V^{2} $
