Black Hole Surface Area-Mass Equation Lets suppose we have a BH with mass $M$ and surface area $A$. 
From the Schwarzschild raidus we can say that the Area of the BH is proportional to its mass
$$A=4\pi r^2=\frac {16 \pi G^2M^2} {c^4}$$ or simply let us say
$$A(M)=\alpha M^2$$
Now let us suppose this black hole gains mass $m$ per second. We can easily see that the gained mass will be proportional to the surface Area of the Black hole. Hence we can write
$$m\propto A$$ 
or $$m(A)=\beta A$$ where $\beta$ is constant.
Now I want to find that the gained mass by the black hole per second ($M(t)=?$).
Lets think like this at $T$ The surface area is $$A=\alpha M^2$$;
$$T:A=\alpha M^2$$
then it gaines mass by amount of $dm$ and it increases the area by $dA$. 
So at $$T+dt:A+dA=\alpha (M+dm)^2$$
And the gained mass at this $dt$ time will be equal to $dm=dA\beta$ or can we say $dm=A\beta$
Edit: Lets thing mass $m$ around the surface of the black hole. The mass has a density $p_m$. By using this we can write such equation. 
$dm=Ap_mdr$ or $m=\int Ap_mdr=\int 4\pi r^2dr$ for $r>R$ where $R$ is the radius of the black hole. I was thinking that the $dM/dt=m$ but I thought later that units dont match. But I cannot stop thinking that the logical answer should be like this since the "change in the black hole mass" should be equal to the gained mass right ? 
 A: The quantity $A$ is actually the surface area of the event horizon of the black hole, as seen from a faraway, stationary observer; as such, it may not behave like the quantity that we traditionally call surface area.
If we differentiate
$$A=\frac{16\pi G^2 M^2}{c^4}$$
with respect to time, we get
$$\frac{dA}{dt}=\frac{32\pi G^2 M}{c^4} \frac{dM}{dt}$$
which tells us that the rate at which the surface area of the event horizon grows, $\frac{dA}{dt}$, is dependent on both the current mass $M$ of the black hole and the rate at which the black hole accretes mass, $\frac{dM}{dt}$. If you know the rate at which the surface area of the event horizon grows, and you know the mass of the black hole, then you can determine how much mass the black hole is accreting. If you do not know the rate at which the surface area of the event horizon is growing, then you also do not know the rate at which the mass is growing.
As to your claim that the rate of mass gain is proportional to the surface area of the event horizon, that is incorrect in several ways. To see why, let's rearrange the first equation:
$$M=\frac{c^2}{4G}\sqrt{\frac{A}{\pi}}$$
and differentiate with respect to time:
$$\frac{dM}{dt}=\frac{c^2}{8G}\sqrt{\frac{1}{A\pi}}\frac{dA}{dt}$$
We can see, firstly, that your assertion is incomplete; the rate of mass gain is dependent on both the area of the event horizon and the rate at which that area is growing, neither of which are necessarily constant. Secondly, your assertion that the rate of mass gain is proportional to the area of the event horizon is incorrect. Rather, it's proportional to $\frac{1}{\sqrt{A}}$. If we say that the rate of mass gain is $m$ and the rate that the area is growing is $a$, then we can fix your statement in the following way:
$$m(A,a)=\beta\frac{a}{\sqrt{A}}$$
If you want to know $M(t)$, though, all of this isn't really helping you. All you need to find $M(t)$ is $A(t)$. Plug $A(t)$ into the first equation and you have $M(t)$.
