I don't know offhand but you can figure it out. You cannot use any voltage you want. The voltage will determine the max rated output power. For a particular load resistance, there is only one supply voltage that will give a particular desired max rated output power level.

Take the transformer secondary's rated RMS output voltage times sqrt(2), i.e. x 1.414, then subtract 1.5 V or so for rectifiers, and that's your nominal rail voltage, **Vrail**, when under load.

Then subtract the "clipping voltage" from the graph in the LM1875 datasheet (for your loaded-down rail voltage value, Vrail). If there is no graph, use 2.5 to 3 Volts for **Vclipping**.

Then subtract 10% of Vrail, for the ripple voltage, **Vripple**. Call the result **Vpeak**, since it will be the peak signal voltage you can have, when running at max output power.

Then peak output power = Vpeak x Vpeak / Rload. And the maximum continuous RMS output power will be (Vpeak / 1.414 ) x (Vpeak / 1.414 ) / Rload.

For the reservoir capacitance, use

C = [ (1000000 * Vpeak) / (Rload x (Vripple) ) ] * [ (1/(2 * fmains)) + (0.02 / Cap_voltage_rating) ] microfarads,

where Vripple = Vrail - Vclipping - Vpeak, or, Vripple = 0.10 x Vrail .

Since an amplifier is usually only a 1/2 to 2/3 efficient, typically 55% to 65%, use a **transformer VA rating** that is at least 2X the max RMS output power that you calculated (actually, use 4X, since you calculated for only one channel).