When you full-wave-rectify the mains voltage, the cap has to deliver all the load current for up to 10msec. According to the "thumb" formula

U = Q / C = I x t / C = 10A x 10msec / 3600µF = 27.8V

a maximum load current of 10A results in a voltage drop of about 28V.

Even if the rectifier is able to fully charge up the storage cap afterwards, periodically every 10msec, then you will still observe a ripple of about 28Vpp and an average output voltage of about 300V - 28V / 2 = 286Vdc. This is the maximum average output voltage you can expect, if you would use an ideal rectifier, an ideal storage cap (of 3600µF) and an ideal mains voltage supply.

But, real world isn't ideal: Fully charging up the storage cap periodically every 10msec means, that an enormous current must flow during a very brief period of time: If the charging period lasts about 1msec, then about 10 times the load current must flow, means roughly 100A!

Can your rectifier withstand 100A for a period of 1msec, periodically every 10msec? If not, then much higher voltage drops must be expected than your given 1.6V! Not even mentioning, that the rectifier will become damaged rather quickly...

Also, 100A will cause considerable voltage drop across all involved connections and soldering joints.

Often overlooked is the equivalent series resistance of storage cap: A typical 3300µF/450V electrolytic shows about 40mOhm. When charging this cap by a current of 100A, 4V will drop across the internal series resistance. This voltage will not appear as storage voltage later, but only heat up the cap, by 400W, by the way. These 4V are lost, your storage cap will show a voltage, which is 4V smaller than when using an ideal cap!

So, I think your problem is, that you suffer from many hidden voltage drops, adding and adding and adding.

As others have already mentioned, such a supply drawing very high currents from the mains supply during brief periods of time will probably violate the CE standards. A power factor correcter might be needed.

Kai