According to the sources, for an ideal transformer, the power input to the primary is equal to the power output from the secondary ($P_p = P_s$) . The power consumed by the lamp in the secondary circuit is given as 60 W. Therefore, the power in the primary circuit is also 60 W. Using the power formula $P = V I$, we can find the primary current ($I_p$) by dividing the primary power by the primary voltage: $I_p = P_p / V_p = 60 \text{ W} / 220 \text{ V} \approx 0.2727$ A, which rounds to 0.27 A.