What is the magnitude of δv12 if the bulb is removed from the socket (i. e. the circuit is not closed) ?

Since the circuit is incomplete or not closed, no current flows in the circuit. as per ohm's law, Voltage is directly proportional to current and is given as

V = Voltage = i R where i = current, R = resistance

as no current flows in the circuit, i = 0

the resistance R can not be zero. hence

V = 0 (R)

V = 0 Volts

so the magnitude of the Voltage is zero Volts

If the socket is open after removing the bulb, then the substitute value is

I = 0 A

ΔV₁₂ = IR

= (OA) (2Ω)

= 0 V