To small numerical factors, the fusion reaction rate is $r_{AB} \propto n_A n_B <\sigma_{AB}\, v>$, where $<\sigma_{AB}\, v>$ is the (temperature-dependent) "reactivity" for the reaction, formed from the averaging the cross-section over an appropriate Maxwellian velocity distribution, and $n$ are the number densities of the reactants.
The proton density at the centre of the Sun is about $10^{32}$ m$^{-3}$, the temperature is $1.5\times 10^{7}$ K. The initial (rate-determining) step in the p-p chain is the formation and subsequent beta decay of a di-proton. The cross-section for this reaction is $\sim 10^{-23}$ barns and $<\sigma_{pp}\, v> \sim 10^{-43}$ cm$^{3}$ s$^{-1}$ at the solar core temperature. At the solar core, this reaction rate yields just 250 W m$^{-3}$.
For the deuterium-tritium reaction used in controlled fusion experiments on Earth, the temperatures are $\sim 10^{8}$ K and the number densities of the reactants $\sim 10^{20}$ m$^{-3}$ (roughly correct for the JET and ITER reactors). The cross-section of the reaction at this temperature is a few barns, much higher than the p-p reaction, and the reactivity is $<\sigma_{DT}\, v> \sim 10^{-15}$ cm$^3$ s$^{-1}$.
Putting these order of magnitude estimates together, the ratio of reaction rates multiplied by the energy released per reaction is
$$ \frac{r_{DT} \times Q_{DT}}{r_{pp} \times Q_{pp}} \sim \frac{10^{40}\ 10^{-15}}{10^{64}\ 10^{-43}} \times \frac{18\ \mathrm{MeV}}{26\ \mathrm{MeV}} = 10^{4}$$
In other words, controlled nuclear fusion experiments (briefly) yield roughly $2.5 \times 10^{6}$ W m$^{-3}$ Thus the energy released per unit volume and the reaction rate per unit volume are about 4 orders of magnitude larger/faster in an Earth-based fusion reactor compared to the core of the Sun.