Self Inductance Of the Coil
Let's review the formulas that apply:

Weber-Turns (flux linkage):
$$ \lambda=N\Phi $$

Flux:
$$ \Phi=\frac {Ni} {\Re}$$

where \$\Re\$ is the reluctance, that depends on the coil's core material and geometry.

Voltage (emf)

$$ V=\frac {d\lambda}{dt}=\frac {N^2} {\Re} \frac {di}{dt}=L\frac {di}{dt}$$

The problem here is that it is not specified how the coil is excited. 

Coil excited by a sinewave current

If it is excited by a sinewave current, then it is rather clear that lower inductance, implies lower flux linkage, lower emf. About the flux it is given by \$\Phi=\frac {Ni} {\Re}=\frac L N i\$. So reducing L alone doesn't give an indication of what \$\Phi\$ will do. For example, N can be reduced in more proportion than L if one choose a different coil material for example. Therefore \$\Phi\$ can be bigger or lesser than before.

Coil excited by a voltage

In contrast, if the coil excited by a voltage, then the flux linkage is given by the voltage alone (because \$V=\frac {d\lambda}{dt}\$, linkage is the voltage integration), and so doesn't depends on the inductance. Also the flux doesn't depends really on the inductance in this case, but just on the number of turns. (And because if wished you can have a big inductance with few turns by selecting a different core, then you might have a big or smaller flux by reducing L). Finally, again if excited by a voltage, the emf will be equal to it, so doesn't depends on the inductance.

Actually whatever excitation you consider, the first 3 options are wrong. 

The last option asks about a steady current, so one can assume it is excited by a voltage in series with a resistor.