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.
If self inductance is infinite, then any rate of change of flux would generate infinite voltage, but look at it the other way. Any amount of back emf can be generated with infinitesimally small rate of change of flux. When we apply a voltage to the primary of a transformer (with no secondary load), an equal and opposite back emf is initially generated and no current flows. In order to maintain the back emf, there must be a rate of change of flux, which requires an increasing primary current. The larger the self inductance, the smaller the rate of current rise required to maintain the emf. As self inductance approaches infinity the rate of rise of current approaches zero. So in the ideal transformer one could apply a voltage to the primary for ever with no current flow. The voltage across the socondary will be the primary voltage times the turns ratio. The current in the primary will be the secondary current / turns ratio.
Best Answer
Let's start by defining the two terms:
Self-inductance is a measure or coefficient of self-induction in a circuit, usually measured in henries. It is also the property of an electric circuit that permits self-induction.
Leakage inductance is an inductive component present in a transformer that results from the imperfect magnetic linking of one winding to another.
Here is a longer definition of leakage inductance:
It also might help you if you first understand what inductance is:
http://www.youtube.com/watch?v=84kl_WheNtQ
Here is another which is also very helpful:
http://www.youtube.com/watch?v=NgwXkUt3XxQ
Essentially, inductance is when the energy is stored in an electromagnetic field.
Now that you grasp inductance, let's look at self-inductance.
Self-inductance is what happens when the current can change in a coil as a result of itself. It occurs as a result of the electromagnetic field (EMF), and has a relation to the proximity and shape of the coil predicting the scope and depth of the EMF.
Here is a video explaining self-inductance:
http://www.youtube.com/watch?v=LwwSJhEfJeA
Here is another video on self-inductance:
http://www.youtube.com/watch?v=B8CPGiK59f8
Even if you don't know the math, you should be able to glean the basic concept.
In a coil, the electromagnetic fields overlap. As a result, the overlapping EMF can feed itself (to an extent), and this is called self-inductance.
Leakage inductance, on the other hand, is a problem, not a property. Leakage inductance happens when self-inductance occurs undesirably, and by definition, leakage inductance will result in an unwanted outcome unless it is mitigated.
As a result of leakage inductance, what will happen is that the magnetic field does not follow the intended path. This will disrupt the flow of the EMF unless addressed.
Whereas self-inductance is a property of leakage inductance, leakage inductance is not a property of self-inductance.