Let us use Ohm's law and rated Voltage and Current of the load to calculate its resistance #R_L=120/6=20Omega#

Let #R_W# be resistance of requisite wire.

Total resistance#=R_L+R_W#

Current in circuit#=120/(20+R_W)A#

Voltage drop across wire#=R_Wxx120/(20+R_W)V#

Equating to the given value we get

#5=R_Wxx120/(20+R_W)#

#=>(24R_W)/(20+R_W)=1#

#=>(24R_W)=(20+R_W)#

#=>R_W=20/23Omega# .....(1)

The electrical resistance of a wire is dependent on its length #L#, its area of cross section #A#, and upon the material of wire accounted through #rho# resistivity of material. Resistance of a wire can be expressed as

#R=(rhoxxL)/A# ......(2)

It is also temperature dependent. Resistivity of copper at #20^@ "C"# is #1.724xx10^-8 Omegam#. Equating (1) and (2) and inserting given values in SI units we get

#20/23=(1.724xx10^-8xx(200xx0.3048))/A#

#=>A=(1.724xx10^-8xx(200xx0.3048)xx23)/20#

#=>A=1.2xx10^-6m^2#

From (1) we also have

#R_"Wire"=20/23xx1000/200=4.35Omega" per 1000 ft"#

This corresponds to U.S. wire gauge of #16AWG# which has a resistance of #4.016Omega" per 1000 ft at "20^@"C"#