What will be the intensity of a lamp if the voltage is 220v and the resistance 42?

You must apply Ohm's law I = V / R, like this: 220/42 = 5.24 Amps. If the lamp is short-circuited R = 0 Ohms, so, in theory, the current would be infinite, obviously this will not be the case, because the protection fuse will jump first.

As well says Joaquin, the initial intensity will be of 5.24 but if you factor in the power of the bulb: By 5.24 x 220V =1152,8 W If you look at the bulb you'll see that it doesn't have that power, not by a long shot, it is because the resistance this cold and lets in more current when they warm up, increases its resistance and causes it to pass less current. In accordance with the consumption marked by the light bulb. It is important to understand that a light-bulb turning on can consume 10 times more of what it consumes once it has warmed up (few seconds after)