Hi, Paresh Gaikar,
At first, we can simplify the specific capacitance into units of current and time, as shown in the equation below, with voltage present in the denominator part of the equation.
C / A = q / V.A = I.t / V.A units: [A.s/V.cm2] = [mA.h / 3.6 . V . cm2]
Therefore, If we multiply the specific capacitance by potential (V) we have the following equation, that may represent the specific capacity (lets name it as X here):
X = C.V / A = q.V / V.A = I.t/A = q / A units: [A.s / cm2] = [mA.h / 3.6 cm2]
With the units you were looking for. Therefore, the specific capacity represents the stored charges accumulated in the capacitor's plates per unit of area.
I hope it may help you.
Boratto is right. Just multiply your value in F/cm-2 by (V/3.6) to get in mAh/cm-2
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