Answer
(a) bulb A
(b) 4
Work Step by Step
(a) We know that power and voltage are related as
$P=VI$
$\implies P_A=4P_B$
$\implies VI_A=4VI_B$
$\implies I_A=4I_B$
Thus, the bulb A has the greater current passing through it.
(b) We know that power and voltage are related as
$P=VI$
$\implies P_A=4P_B$
$\implies VI_A=4VI_B$
$\implies I_A=4I_B$
$\implies \frac{I_A}{I_B}=4 $
This is the required ratio.