Three 9w -3 v bulbs is to be lighted by a suppy of 12 volt. for this, a wire is to be connected in series with it. th resistance of the wire should be:
a) 1
b) 2
c) 3
d) 4

9W 3V bulb need current flow
I = P/V = 9/3 = 3A

connected serially, voltage drop on the wire will be = Vsupply - 3(Vbulb) = 12 - 3(3) = 3V.

resistance of the wire ,with I=3 and V=3, = V/I = 3/3 = 1 Ohm.

In order for a 3V bulb to dissipate 9W, the current through it must be 3A.
R = E/I = 3v/3A .  The resistance of such a bulb is 1 ohm.

There are two correct answers to the question, since the question itself
is ambiguous and may be understood in two different ways (at least) ...
series and parallel.

If the three bulbs are in series and the mystery wire will be placed
in series with them, then . . . . .

-- The current everywhere in the series circuit is 3 A .
-- The resistance across the supply is  R = E/I = 12v/3A  =  4 ohms.
-- 3 of the total 4 ohms is in the bulbs.
-- The additional wire must provide the additional 1 ohm .

If the three bulbs are in parallel and the mystery wire will be placed
in series with them, then . . . . .

-- The three bulbs in parallel form an effective resistance of 1/3 ohm.
-- The voltage across the parallel section of the circuit is 3v.
-- The additional wire must drop 9 volts across itself. 
-- The total current through the 3 branches of the parallel section,
    as well as through the additional piece of wire in series with it, is 9 A.
--  The resistance of the wire must be  R = E/I  =  9v/9A = 1 ohm.

Verrry interesting ! 
Whether the bulbs are in series or in parallel,
the same 1-ohm piece of wire does the job.

Are there options for this?