The configuration of the solar system is thought to have been more or less settled after the first 10-20 million years or so. However, what governs the Earth's orbital period is it's orbital angular momentum and the mass of the Sun. Two events have certainly changed the Earth's orbital period (a) whatever collision formed the Moon and (b) the continuous process of mass loss from the Sun.
Given that (a) probably happened sometime in the first tens of millions of years and likely did not alter the Earth's angular momentum greatly - it depends on the speed, mass and direction of the impactor and the amount of mass lost from the Earth-Moon system - I will ignore it.
We know a little more about (b). It seems that the mass loss from the early Sun was much greater than it is now. A review by Guedel (2007) suggests a mass loss rate over the last 4.5 billion years that increases as $t^{-2.3}$ (with considerable uncertainty on the power law index), where $t$ is time since birth, and an initial solar mass between 1% and 7% more than it is now.
Conservation of angular momentum and Kepler's third law means that $a \propto M^{-1}$ and $P \propto M^{-2}$. Therefore the Earth's orbital period was 2-14% shorter in the past.
What this considerable uncertainty means is that your question cannot be answered to three significant figures, or perhaps even two.
If the solar wind power law time dependence is very steep, then most of the mass loss occurred early, but the total mass loss would have been greater. On the other hand, a lower total mass loss implies a shallower mass loss and the earth spending a longer time in smaller orbit.
If I have time I will run through a calculation, but it seems safe to assume that the earth has executed more orbits than its age in years, but probably not by more than a few percent.
EDIT: A further consideration could be the tidal torque exerted by the Sun on the Earth-Sun orbit, which would increase the orbital separation.
Quantifying this is difficult. The tidal torque on a planet from a the Sun is $$ T = \frac{3}{2} \frac{k_E}{Q} \frac{GM_{\odot}^{2} R_{E}^{5}}{a^6},$$ where $R_E$ is an Earth radius and $k_E/Q$ is the ratio of the tidal Love number and $Q$ a tidal dissipation factor (see Sasaki et al. (2012).
These lecture notes suggest values of $k_E/Q\sim 0.1$ for the Earth and therefore a tidal torque of $4\times 10^{16}$ Nm. Given that the orbital angular momentum of the Earth is $\sim 3\times 10^{40}$ kgm$^2$s$^{-1}$, then the timescale to change the Earth's angular momentum (and therefore $a$ and $P$) is $>10^{16}$ years and thus this effect is negligible.