More turns increases inductance and packing the turns closer together (so that they all share each other's magnetic field) maximizes inductance for a given number of turns. Using a ferromagnetic core helps all the turns of the wire share each other's magnetic flux. If the ferromagnetic core becomes a full loop - then the inductance increases dramatically as the "gap" reduces to zero.
Ultimately, on an ungapped core inductance is proportional to turns-squared. It's also proportional to the magnetic permeability of the material.
If the ferromagnetic core is conductive this can dramatically reduce inductance at AC frequencies because the core acts as a shorted turn - this is why power transformers have cores made of insulated laminates of iron.
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.
Best Answer
There is a web-calculator for this. And here, you can find an excellent paper about methods of calculation for different shapes of coils.