To find the current in the primary winding of an ideal step-down transformer, we first understand that an ideal transformer's power input (primary side) equals its power output (secondary side). This is because an ideal transformer is assumed to have 100% efficiency, meaning there are no losses in the transformation process.

The power of the lamp (which is connected to the secondary side of the transformer) is given as $60\,\mathrm{W}$, and it operates at $12\,\mathrm{V}$. The voltage on the primary side of the transformer is $220\,\mathrm{V}$.

Using the power formula $P = IV$, where $P$ is the power in watts, $I$ is the current in amperes, and $V$ is the voltage in volts, we equate the power on both sides of the transformer because of its ideal nature, i.e., $$P_{\text{primary}} = P_{\text{secondary}}$$.

Hence, $$60\,\mathrm{W} = 220\,\mathrm{V} \times I_{\text{primary}}$$.

Solving for $I_{\text{primary}}$ gives:

$$I_{\text{primary}} = \frac{60\,\mathrm{W}}{220\,\mathrm{V}} = 0.2727... \approx 0.27\,\mathrm{A}$$.

Therefore, the current in the primary winding of the transformer is approximately $0.27\,\mathrm{A}$.