The deuterium-tritium fusion reaction cross-section is highly temperature dependent and peaks at temperature of about $8\times 10^{8}$ K, so I suppose these are the temperatures to aim for in a controlled nuclear fusion experiment. In fact according to this, the operating temperatures are at least $10^{8}$ K.
The density of the fusion plasma is a factor - the reaction rate will be proportional to the product of the densities of the two reactants. In fusion reactors the density is of order $10^{20}$ m$^{-3}$. At the centre of the Sun the particle densities are 12 orders of magnitude higher, so partly the increased temperatures in a fusion reactor are to compensate for the lower densities. However, it is also worth remembering that the Sun is not a particularly intense fusion reactor. It only produces about 250 W per cubic metre in its core. A bigger compensatory factor is that the Deuterium-tritium fusion cross-section is about 25 orders orders of magnitude greater than that for proton-proton fusion in the Sun.
In this question I have posted an answer that estimates the energy release per unit volume in typical reactor conditions versus the Sun. I find (order of magnitude) that you get $10^{4}$ times more energy per unit volume out of a reactor than the core of the Sun. So $\sim 10^{6}$ W m$^{-3}$, which I guess is what you will need to make it commercially viable. If you dropped the temperature at all it would rapidly become unviable as a significant power source without absolutely enormous reactors.