Dry air will break down if the electric field exceeds about 3.0 ã 106 v/m. what amount of charge can be placed on a capacitor if the area of each plate is 7.9 cm2? \

Although this question does not say so explicitly, the first sentence suggests that it is about dielectric breakdown in a parallel-plate air-dielectric capacitor (I am making most of this up, since actually the question says nothing about the form or dielectric of the capacitor). 

It seems that we are being asked to use Q = V*C which relates voltage (V), charge (Q) and capacitance (C) together with C = A*e*e0/t (A= area of parallel-plate cap, e = dielectric constant of the dielectric, e = permittivity of free space, t = separation of plates), and also E = V/t (E = electric field strength). 

Put all of this together and you can get:- 

E = V/t = Q/C*t = Q*t/(A*e*e0*t) = Q/(A*e*e0) 

or Q = E*A*e*e0 

If we set E to the dielectric strength of the dielectric, this equation sets a maximum value for the charge which can be stored on the cap. If you are prepared to make the assumptions I have made (i.e that the question is intended to refer to a parallel plate cap with air dielectric) then you can answer it by substituting A = 54*10^-4m^2, E = 3*10^6V/m and e = ~1 and solving for Q.

Independent variable is something that is changed in a experiment.