The solar constant is $S=L_\odot/4\pi a^2$ where $L_\odot$ is luminosity and $a$ is Earth's orbital semimajor axis. So if you want to reduce it by $\Delta S$ you need to increase the semimajor axis to $$a'=\frac{a}{\sqrt{1-\Delta S/S}}.$$
Lifting earth straight from $a$ to $a'$ would take $GM_\odot M_\oplus(1/a-1/a')$ energy, but presumably we also want it to remain in a nice circular orbit so we need to adjust the kinetic energy to match.
The total energy of Earth is $$E=(1/2)M_\oplus v^2 - \frac{GM_\oplus M_\odot}{a}$$ where $v=\sqrt{GM_\odot/a}$ (from the vis-viva equation for a circular orbit). Plugging in, we get $-(1/2)GM_\odot M_\oplus/a$. So moving it from $a$ to $a'$ while adjusting the velocity downward takes $$\Delta E = (1/2)GM_\odot M_\oplus\left(\frac{1}{a}-\frac{1}{a'}\right).$$ It is actually a bit cheaper than a straight lift... assuming we can convert the kinetic energy into potential energy in a safe way.
So if the solar constant change is small $\Delta S \ll S$ so it is nearly constant the time to get lift is $$ t =(1/2)\left(\frac{1}{\pi R^2_\oplus S}\right)GM_\oplus M_\odot \left(\frac{1}{a} - \frac{1}{a'}\right) = (1/2)\left(\frac{1}{\pi R^2_\oplus S}\right)GM_\oplus M_\odot \left(\frac{1 - \sqrt{1-\Delta S/S}}{a}\right).$$
In reality there will be a bit less solar input later on, so this will be a mild underestimate. Inefficiencies in your orbital mechanics will move it towards the double, corresponding to just lifting the planet.
So, plugging in, I get the time to reduce the solar constant by 1 W/m^2 to be $\approx 175000$ years.
If we are escaping a brightening sun one would want to keep the solar constant constant, of course. So the actual rate of expansion would be significantly slower.