no-load DC output voltage of DC wall-wart circuit
Your formula is ignoring the voltage drop from the diodes (given as 0.7V each). Otherwise it it is correct. To derive the sqrt(2) part you need to know that transformer output voltages are usually given as Vrms (volts root-mean-square). The problem doesn't explicitly say that but it is implied. You should be able to search that term and find a good description of the math involved.

Once you have the correct equation you will see that changing the diode voltage will change the output voltage. If the voltage drop was increased to 1.1V you would lose a total of 2.2V in the bridge rectifier, since the current always has to flow through two diodes.