This Khan Academy Video explains it very nicely.

So since this is a first order reaction $$\ce{ -\Delta N2O5/\Delta t = k[N2O5]}$$

This gives you the rate of reaction at the very start. A second later however, $\ce{[N205]}$ decreases which will in turn change the rate of reaction. So lets apply calculus. We can write this as

$$\ce{ -d N2O5/d t = k[N2O5]}$$

Then we can integrate to find out the total change over a period of time.

$$\int_{[N2O5]_0}^{[N2O5]t} d[N2O5]/[N2O5]= \int_0^t-k dt$$

since -k is constant, we can take it outside of teh integral

$$\int_{[N2O5]_0}^{[N2O5]t} d[N2O5]/[N2O5]= -k\int_0^tdt$$

So then we get

$$ ln[N2O5]_t - ln[N2O5]_0 =-kt$$

So putting in your values

$$ln(0.20e-2) - ln(1.24e-2) = -k*60*60$$

So I'm pretty sure the answer is $$5.068e-4\frac{1}{s}$$

This Khan Academy Video explains it very nicely.

So since this is a first order reaction $$\ce{ -\Delta N2O5/\Delta t = k[N2O5]}$$

This gives you the rate of reaction at the very start. A second later however, $\ce{[N2 O5]}$ decreases which will in turn change the rate of reaction. So lets apply calculus. We can write this as

$$\ce{ \frac{-d [N2O5]}{d t} = k[N2O5]}$$

Then we can integrate to find out the total change over a period of time.

$$\int_{[N2O5]_0}^{[N2O5]t} \frac{d[N2O5]}{[N2O5]}= \int_0^t-k dt$$

since $-k$ is constant, we can take it outside of the integral

$$\int_{[N2O5]_0}^{[N2O5]t} \frac{d[N2O5]}{[N2O5]}= -k\int_0^tdt$$

So then we get

$$ \ln([N2O5]_t) - \ln([N2O5]_0) =-kt$$

So putting in your values

$$\ln(0.20\cdot 10^{-2}) - \ln(1.24\cdot 10^{-2}) = -k\cdot 60\cdot 60$$

So I'm pretty sure the answer is $$5.068\cdot10^{-4}~\pu{s^{-1}}$$