60/5 A CT Burden Resistance
As you say, you can simply apply Ohm's law, which gives you 80m\$\Omega\$ for 400mV. The power dissipation at 5A secondary current will be 2W which is fairly reasonable, but it's pretty much irrelevant. The output current is fixed (so long as you stay within the acceptable range for the particular CT) so if you want V volts the resistance cannot vary from that number. 

Of course you are free to use a lower value resistance (say 20m\$\Omega\$) which would give you 100mV and only dissipate 0.5W maximum, but would only give 100mV RMS full scale, so you might need an amplifier. The use of a 4-wire (Kelvin) connection becomes even more important at such low resistances as only 200\$\mu\Omega\$ in the wrong place will represent a relatively large 1% error.