How small a star can provide Sun-level illumination to its planets?

What you need is a mass-luminosity relation combined with an expression for the tidal radius in terms of the stellar mass. The latter also depends on the mass of the planet, so when you say earth-sized, I'll assume that means mass and radius.

So going through the calculation.

Flux at the planet is $L/4pi r^2$, where $L$ is the luminosity and $r$ the orbital radius (assumed circular).

Let's next use an approximate relation that
$$frac{L}{L_{odot}} = left(frac{M}{M_{odot}}right)^3$$
A more accurate numerical relationship could be obtained from evolutionary model calculations. There is a complication that the luminosity of very low mass stars and brown dwarfs does depend on age.

For a rigid body the tidal radius (Roche limit) is
$$ r_tsimeq 1.4 R_Eleft(frac{M_{odot}}{M_E}right)^{1/3},$$
where $R_E$ and $M_E$ are the radius and mass of the Earth.

So setting $r=r_t$, the flux to $1400$ W m$^{-2}$ and replacing $L$ in terms of mass, we find
$$frac{M}{M_{odot}} = left[frac{4pi times 1400}{L_{odot}} (1.4 R_E)^2 left(frac{M_{odot}}{M_E}right)^{2/3}right]^{1/3}.$$

All that remains is to put the numbers in and I get $M = 0.026 M_{odot}$.

So, a small brown dwarf - but there are caveats. First, and most importantly, as I said, the luminosity-mass relation is a bit rough and ready, and is certainly age dependent for something as low mass as this. Second, the expression for the Roche limit depends a bit on structural properties of the planet, but I think this introduces relatively little uncertainty.

EDIT: If you use that mass and the tidal radius formula, you find that the orbital radius is only $5R_E$. As the minimum size of a brown dwarf is about a Jupiter radius, it appears that tidal breakup would not be the limiting factor for an Earth-like planet.