Is it resistance of filament or current thorugh it which determines brightness of bulb?

Three factors:

1. Power dissipated through resistance (as other answers have noted).

2. Portion of that power that’s lost as heat vs EM waves.

3. Portion of the portion lost as EM waves that is visible light vs EM waves outside the visual spectrum.

Resistance and current can only tell us about power ($I^2R$), so we would need more information.

Brightness of a bulb depends on the power it consumes.the more power it consumes the brighter it should glow .

Now power dissipated depends on both current and resistance . If you apply equal voltages then Power = V²/R

It's the power dissipated by the Joule heating of the element that determines the brightness - remember that an incandescent bulb operates on the principle of thermal radiation.

$$P = frac{V^2}{R}$$

so that, in turn, depends on the voltage and the resistance of the element. Most likely it's a bit more complicated than that since the resistance will go up as it heats up. But this is good enough for the general idea. A more resistive bulb will dissipate less power, and so be dimmer, for a given applied voltage, and increasing the voltage dramatically increases the power, as can be see by the square factor. So the less-resisting bulb is going to be the brighter one if you apply the same voltage to both. For current, it's

$$P = IV$$

so goes up linearly with increase in either one, but current is related to voltage with resistance, so you in practice don't modulate voltage and current independently (unless you could somehow modulate the resistance of the filament!).