The correct option is B Both B1 and B2 will go out after sometime
When the current is cut off in a L−R circuit, because of induction, current flows for some more time.
When the battery is removed from an L−R circuit, the current decays exponentially. I(t)=ϵRe−t/τ where τ=L/R, the time constant. (In fact the circuit should be closed first and then the battery cut off). Here both the bulbs will go after some time, because the current continues in the circuit for some more time.