For a ideal transformer, current scales inversely proportional to voltage. Put another way, current x voltage is constant. That's because current x voltage is power, and for a ideal transformer, power in and power out are the same.
Of course no transformer is ideal. There will always be some loss, so power in will be a bit higher than power out, with the remainder ending up heating the transformer.
The easiest approach is to start with assuming ideal transformer, then derate according to the efficiency you think you will actually attain. 30 A at 30 V out is 900 VA. 900 VA / 220 V = 4.1 A. At 80% efficiency, for example, that would be (4.1 A)/80% = 5.1 A. 5 A would be cutting it close, but possible with a good transformer. Unless it's a hardship, I'd plan for 6 A input capability at least.
Another factor is inrush current. This can be large when the core is not magnetized, or even magnetized in the opposite direction when power is turned on at the peak of the waveform. This is a real issue, especially in toroidal transformers, which are common for these kinds of applications. The inrush only lasts up to 1/2 cycle, but can be several times larger than the normal operating current. I once blew a 15 A breaker with a toroidal transformer that didn't even have a load connected to it.