Question

At 3-db frequencies current in the series rlc circuit equal current at resonance multiplied by

Answer 1

⛤Hey There⛤
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dB = 20 log i2/i1 
-3dB = 20 (log i2/i1) 
log i2/i1 = -3/20 
i2/i1 = anti log -3/20 
i2/i1 = .708 
i2 = .708 (i1) 
Therefore at the -3.0 dB frequency the current is .708 of what it is at resonance 
I realize this is a trivial difference from that of the the other responders but maybe it will help you to understand how the 3 dB is derived in the first place. The other responders answers (.707) are correct for a -3.01dB difference. 

Note; The half power point expressed in dB is actually equal to; 
10 log P2/P1 = 10 log .5/1 = -3.01dB and therefore; 
-3.01dB = 20 log i2/i1 
i2/i1 = anti log -3.01/20 
i2/i1 = .707 
i2 = .707( i1) 
Therefore at the the -3.01 dB frequency the current is .707 of what it is at resonance.