The current sensitivity of a galvanometer which is used as an ammeter is doubled by doubling the number of turns in the coil . then , the voltage sensitivity of the galvanometer if it is used as voltmeter in the same way would :

remain unchanged 

The correct answer is option 3: remain unchanged 

• Current sensitivity ∝ number of turns (n)
  So, doubling n ⇒ current sensitivity doubles
• Voltage sensitivity = current sensitivity / resistance

But, when number of turns is doubled: Resistance of the coil also doubles (since length of wire increases)

So, 
Voltage sensitivity ∝ n / R
= 2n / 2R = same

Therefore, voltage sensitivity remains unchanged.