As dmckee said, the problem is complex. Fortunately, there are some simplifications that can be made without much damage to the accuracy. I hope I can present them in an accesible manner.
My guess is that since You have constant inlet and ambient temperatures and a target outlet temperature, what You are looking for are the pipe's dimensions and properties (type of material).
At the beginning You can neglect the radiation from the surface - at 500K it barely exists.
As it was mentioned, the easiest way to solve this problem is to compute an algorithm using software like MatLab or Wolfram Mathematica. Also get familiar with dimensionless numbers used in heat transfer theory. Two of them are more important than others - Nusselt number and Reynolds number. Nusselt number is vital for determining forced convective heat transfer coefficients, there are many empirical correlations for Nusselt number available, but they usually depend on the value of Reynolds number. Reynolds number defines whether the flow is laminar, transitional or turbulent. For turbulent flows in horizontal pipe's Gnielinski correlation for convective heat transfer coefficient is quite good. Calculating the heat transfer coefficients You will often have to define the wall temperature. It is generally unknown, it differs not only along the pipe's length but also in the radial direction - it is hotter on the inner surface of the pipe. Since the pipe's walls are usually thin, in Your calculations You can assume that the temperature is uniform along the radius, but You cannot assume that it is uniform along it's length. That's why the easiest way is to solve it iteratively, calculating the fluid's properties after each, let's say 5 cm of the pipe. Each time You can set the wall temperature to
$$\frac {T_{\text{fluid}} + T_{\text{amb}}}{2}$$, which isn't very bad approximation. You must remember though, that $T_{fluid}$ is 500K only at the first step, every next step it should be a result of the previous calculation. You carry on with the calculations unless the temperature of the fluid reaches 330K. Then You count the number of steps, multiply it by 5cm, and the result is the required pipe's length. Of course, at the beginning, pipe's inner and outer diameters shall be assumed as well as the pipe's material (it affects the value of thermal conductivity). Now that You have calculated the length You have to decide if it suits You, and alternatively change the pipe's diameters or material and repeat the calculations until the pipe's length satisfies You.
Although this is a difficult problem to model, a simple lumped thermal analysis can bring some understanding and an approximate solution.
We'll consider the material flow through the pipe to be plug flow and study the temperature of a small mass element $dx$ travelling down the pipe.
Using Newton's cooling/heating law and considering convective heat losses only, we can write:
$$\frac{dQ}{dt}=u(T-T_{\infty})dA,$$
where the LHS is the heat flux leaving the element, $u$ the overall heat transfer coefficient, $dA$ the surface area of the element, $T$ its temperature and $T_{\infty}$ the ambient temperature. Developing a little, we get:
$$\frac{dQ}{dt}=\pi Du(T-T_{\infty})dx,\tag{1}$$
where $D$ is pipe outer diameter.
As the element has lost heat:
$$dQ=-dmc_pdT$$
$c_p$ is the heat capacity of the gas. Dividing both sides by $dt$ and with $\dot{m}=\frac{dm}{dt}$ gives:
$$\frac{dQ}{dt}=-\dot{m}c_pdT,\tag{2}$$
where $\dot{m}$ is the mass throughput of the gas.
Using the identity $(1)=(2)$, we get a simple differential equation:
$$-\dot{m}c_pdT=\pi Du(T-T_{\infty})dx$$
$$\frac{dT}{T-T_{\infty}}=-\frac{\pi Du}{\dot{m}c_p}dx=-\alpha dx,\tag{3}$$
where:
$$\alpha=\frac{\pi Du}{\dot{m}c_p}$$
Integrating $(3)$ between $0, T_1$ and $L, T_2$ gives:
$$\ln\frac{T_2-T_{\infty}}{T_1-T_{\infty}}=-\alpha L,\tag{4}$$
where $T_1$ and $T_2$ are the incoming and outgoing temperatures of the gas, respectively and $L$ is the pipe length. From $(4)$, $T_2$ can easily be extracted.
The overall heat transfer coefficient $u$ can be estimated from:
$$\frac1u\approx\frac{1}{h_1}+\frac{\theta}{k}+\frac{1}{h_2},$$
where $h_1$ is the convection heat transfer coefficient gas/mica, $k$ the thermal conductivity of mica, $\theta$ the wall thickness and $h_2$ is the convection heat transfer coefficient mica/air.
Limitations of the model:
At high temperature, convection is not the only heat loss mode: radiation loss will also be important. This can be rectified by adding a radiative heat loss function to $(1)$. With Stefan-Boltzmann the loss function would be:
$$\sigma\epsilon(T^4-T^4_{\infty})dA$$
Secondly, assuming plug flow in the case of a low speed low viscosity fluid like a hot gass isn't very realistic. Assuming laminar flow requires a far more demanding mathematical approach though.
Numerical evaluation:
$(4)$ reworks to:
$$T_2=T_{\infty}+(T_1-T_{\infty})e^{-\alpha L}$$
To estimate $\alpha$, I used:
$u\approx 17\:\mathrm{Wm^{-1}K^{-1}}$, based on OP data and literature values.
$c_p=1000\:\mathrm{Jkg^{-1}K^{-1}}$
$\dot{m}=0.000009\:\mathrm{kg/s}$
$D=0.010\:\mathrm{m}$
Which gives an estimate of $\alpha\approx 70\:\mathrm{m^{-1}}$.
With $T_1=1100\:\mathrm{C}$ and $T_{\infty}=20\:\mathrm{C}$, with $L=1\:\mathrm{m}$, I get:
$$T_2\approx 20\:\mathrm{C}$$
So over $1\:\mathrm{m}$, the gas would have cooled down completely.
For other pipelengths $x$, evaluate as:
$$T_2=20+1080e^{-70x}$$
Best Answer
Briefly put: you need to know the inlet flow rate of exhaust gasses and the temperature and the power you want to extract. From the power (and your ambient temperature) you also get the flow rate for the water. From this caclulate the log mean temperature difference (LMTD) along the heat exchanger, look for the heat transfer numbers for the configuration your using (one list can be found here - you'll see that there is a wide range of possible transfer values to consider) and you should be able to calculate the surface area you need. A very brief explanation is also found on the wikipedia page for LMTD
Please also note the following: by cooling the exhaust, you may have less natural convection in your chimney, make sure your oven will still work. Also consider a margin of safety in sizing the exchanger, fouling will surely lower the heat transfer over time.
Last not least, this is an engineering question - support engineering stack exchange!