Answer
$(B)\ 0.11s$
Work Step by Step
We know that
$\epsilon=N|\frac{(B_2-B_1)l^2}{\Delta t}|$
This can be rearranged as:
$\Delta t=N|\frac{(B_2-B_1)l^2}{\epsilon}|$
We plug in the known values to obtain:
$\Delta=(4)|\frac{1.9\times 10^{-5}-1.2\times 10^{-5}(0.75)^2}{1.4\times 10^{-4}}|=0.11s$