The lamps are in parallel.
Advantages:
If one lamp is faulty, it will not affect the working of the other lamps. They will also be using the full potential of the battery as they are connected in parallel.
The lamp with the highest power will glow the brightest.
P=VI
In this case, all the bulbs have the same voltage. But lamp C has the highest current.
Hence, for Lamp C
P=5 x 60 Watt = 300 W. (the maximum).
The total current in the circuit = 3+4+5+3 A = 15A
The Voltage = 60V
V=IR and hence R =\(\frac{V}{I}\)
= \(\frac{60}{15}\) A = 4A