$M$ = mass of the Sun
$m$ = mass of the Earth
$r$ = distance between the Earth and the Sun
The sun is converting mass into energy by nuclear fusion.
$$F = \frac{GMm}{r^2} = \frac{mv^2}{r} \rightarrow r = \frac{GM}{v^2}$$
$$\Delta E = \Delta M c^2 = (M_{t} - M_{t+\Delta t}) c^2 \rightarrow \Delta M = \Delta E / c^2$$
$$\rightarrow \frac{\Delta r}{\Delta t} = \frac{G}{v^2 c^2}.\frac{\Delta E}{\Delta t}$$
Sun radiates $3.9 × 10^{26}~\mathrm W = \Delta E/\Delta t$
Velocity of the earth $v = 29.8 \mathrm{km/s}$
There is nothing that is stopping the earth from moving with the same velocity so for centripetal force to balance gravitational force $r$ must change.
Is $r$ increasing? ($\Delta r/ \Delta t = 3.26070717 × 10^{-10} \mathrm{m/s} $)
Answer
I think the reasoning has an error. It assumes $v$ is constant, but instead we ought to assume the angular momentum is constant.
By dimensional analysis that leads to
$$r \propto \frac{L^2}{GM}$$
so as $M$ decreases, $r$ increases (the original post had $r \propto M$, not $r \propto 1/M$.
On the other hand, assuming a circular orbit seems dubious.
As the other commenters said, this effect is minute. A significant effect on the orbit of the moon around the earth is tidal evolution, which does actually push the moon further away.
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