This is a typical example where a Newtonian derivation is much simpler and quicker, and gives the same answer. Which you can easily find online.
But if you want to do this from within GR, then you have to work out the Ricci tensor entry $R_{00}$, the Ricci scalar $R$, and the metric entry $g_{00}$:
$g_{00} = 1$;
$R_{00}$: $$ R_{00} = R^m_{tmt} = R^r_{rtr} + R^\theta_{t\theta t} + R^\phi_{t\phi t} = -3 \frac{\ddot a}{a},$$ where each Riemann tensor depends on the Christoffel symbols (listed for instance in section C here);
$R$: $$R = g^{ik}R_{ik} = -6\frac{\ddot a}{a} - 6\left ( \frac{\dot a}{a} \right )^2 - 6\frac{1}{k^2a^2},$$ where $k^{-2}=0$ for flat space.
So putting it all together: $$ R_{00} -\frac{1}{2}Rg_{00} = -3\frac{\ddot a}{a}+3\frac{\ddot a}{a} + 3\left ( \frac{\dot a}{a} \right )^2.$$
Hence: $$3\left ( \frac{\dot a}{a} \right )^2 = 8\pi G\rho, $$ $$ \Rightarrow \left ( \frac{\dot a}{a} \right )^2 = \frac{8\pi G}{3}\rho. $$