By the way, it may be helpful if you tell to the community more of the physical problem you want to describe.
I added a source of heat at the origin of the square (your q substituted by an e-function) and you can see how the temperature there rises quite fast ( this did not work when I placed it in the middle of the square. I don't know why)
To get nice pics I changed the time domain by dividing your k by a factor of 1000.
ec = rho*cp*D[T[x, y, t], t] ==
k*(D[T[x, y, t], x, x] + D[T[x, y, t], y, y]) +
1000 Exp[-5 (x^2 + y^2^2)];
f = 6780000; e = 30000; \[Epsilon] = 55.33*10^-12;
k = 104/1000; h = 10; rho = 410; cp = 2.2; To = 300; Tamb = 323;
q = f*e^2*\[Epsilon];
alpha = k/(rho*cp); \[CapitalOmega] = Rectangle[{0, 0}, {0.02, 0.01}];
ci = T[x, y, 0] == To;
cc = {D[T[x, y, t], x] == 0 /. x -> 0,
D[T[x, y, t], y] == 0 /. y -> 0, D[T[x, y, t], x] == 0 /. x -> 1,
D[T[x, y, t], y] == 0 /. y -> 1}
sln = NDSolve[{ec, ci, cc}, T, {t, 0, 1000}, {x, 0, 1}, {y, 0, 1}];
fT = T /. Flatten[sln]
and show it
Animate[Plot3D[fT[x, y, t], {x, 0, 1}, {y, 0, 1}, PlotRange -> {300, 400}], {t, 0, 500}]
If this source of heat dies away, here exponentially, the temperature rises as well, this gets slower and the temperature is equilibrated throughout the square.
ec = rho*cp*D[T[x, y, t], t] ==
k*(D[T[x, y, t], x, x] + D[T[x, y, t], y, y]) +
1000 Exp[-5 (x^2 + y^2^2)] Exp[-.01 t];
f = 6780000; e = 30000; \[Epsilon] = 55.33*10^-12;
k = 104/1000; h = 10; rho = 410; cp = 2.2; To = 300; Tamb = 323;
q = f*e^2*\[Epsilon];
alpha = k/(rho*cp); \[CapitalOmega] = Rectangle[{0, 0}, {0.02, 0.01}];
ci = T[x, y, 0] == To;
cc = {D[T[x, y, t], x] == 0 /. x -> 0,
D[T[x, y, t], y] == 0 /. y -> 0, D[T[x, y, t], x] == 0 /. x -> 1,
D[T[x, y, t], y] == 0 /. y -> 1}
sln = NDSolve[{ec, ci, cc}, T, {t, 0, 1000}, {x, 0, 1}, {y, 0, 1}];
fT = T /. Flatten[sln]
Animate[Plot3D[fT[x, y, t], {x, 0, 1}, {y, 0, 1}, PlotRange -> {300, 400}], {t, 0, 1000}]