Hello,

A topology A is said to be finer than a topology B, if any open set of B is an open set of A.

Knowing that, it's quite easy

Consider the less fine topology, let's say {0,{a},X} (0 designs the empty set)

Then you have to find a topology that contains {a} (it'll always contain 0 and X, because it's the definition) and more sets.

So you can add {b} and {a,b}. You have to add {a,b} because the union of two elements of a topology is in the topology and {a} U {b} = {a,b}

Hence {0,{a},{b},{a,b},X} can be the second finest topology.

You need to find a third one, that contains {a},{b},{a,b}

Let's add {c}

Since it has to be stable by union (for intersection, it's trivial because it's the empty set), you have to add {a} U {c}={a,c}, {b} U {c}={b,c} and {a,b} U {c}={a,b,c}

So the finest topology can be {0,{a},{b},{c},{a,b},{a,c},{b,c},{a,b,c},X}