When you are given acceleration, density, area and time, you can indeed find an expression for mass in terms of these. Here is how you go about it:
Make a table of the units that occur in each, and their exponents:
L M T
a 1 -2
D -3 1
As you can see, you need to use D (density) as the only one that contains mass. But when you use it, you bring length along (with an exponent of -3). Now you can simply look up and down your table for a factor that will eliminate length. A will do that.
I will leave it as an exercise for you to see what coefficients A and D need to have in the expression
$m = D^\alpha A^\beta$
Note that you could write the answer as "any set of coefficients that solves" four equations with three unknowns; this means that there is more than one solution - infintely many, in fact. For example, $at^2$ has units of length, so $D\cdot(at^2)^3$ is one of many combinations that yields mass - but the expression I wrote earlier is simpler. And any product of powers of $at^2$ and $A$ also has dimensions of length (to some power) and could therefore be combined with density to give mass.
As Kyle Kanos stated earlier, mass is a fundamental unit; that doesn't mean you can't express it in terms of derived units, although it's slightly unusual to do so - but certainly not unheard of. Dimensional analysis is very powerful - it's well worth getting to know it better.
And if you are actually looking for $\alpha$ and $\beta$ in the expression you gave in your question - that is indeed impossible. Since neither length nor time contain "mass" in their units, there is no way to make mass appear by multiplying them together. That's how "fundamental units" work.