Step 1: Recall the power voltage current relation.
Electrical power obeys \[ P = VI \] so current can be isolated as \[ I = \frac{P}{V} \]
Step 2: Plug in the lamp's rating.
With \(P = 100\) W and \(V = 220\) V, \[ I = \frac{100}{220} \]
Step 3: Simplify the fraction.
Both numbers share a factor of 20, so dividing top and bottom by 20 gives \[ I = \frac{5}{11}\text{ A} \] which is roughly 0.45 A.
\[ \boxed{\frac{5}{11}\text{ A}} \]