Answer
Resistance of each lightbulb: 560Ω
Fraction of power wasted in leads: 0.020
Work Step by Step
The total resistance of the bulb can be found using the equation $R=\frac{V}{I}$
Because the 8 lightbulbs are connected in parallel, the total current is $8\times0.21A=1.68A$ and the total voltage is 120V
$R_{total}=\frac{V}{I}=\frac{120V}{1.68A}=71.4Ω$
The resistance in the 8 lightbulbs is the total resistance minus the resistance in the wires $R_{lightbulbs}=71.4Ω-1.4Ω=70Ω$ and the resistance of one lightbulb is $70Ω\times8=560Ω$
The fraction of power wasted in the leads can be found by
$\frac{P_{leads}}{P_{total}}=\frac{I^2R_{leads}}{I^2R_{total}}=\frac{R_{leads}}{R_{total}}=\frac{1.4Ω}{71.4Ω}=0.020$
