A dibasic organic acid (0.1208 g) requires 23.1 cm3 of 0.1 mol dm-3 NaOH for neutralisation. What is its approximate relative molecular mass?

H2A + 2 NaOH à Na2A + 2 H2O

Another two parter

1st work out the amount of moles of NaOH (see the other post/ the website I posted)

Notice that H2A (that's your acid I'm assuming?) is in a ratio of 1:2
So remember to times the NaOH molarity by two to get the moles of H2A

Part two

The MW (molecular mass) can be worked out using this equation

mass (g) = amount (moles) x molecular weight (g/mole)

just rearrange it to get the MW

so the moles for NaOH is 23.1/40 = 0.5775

0.1M (mol/dm^3) x (23.2cm^3/1000) = 0.00232M

Right I have taken the concentration and times it by the volume -I have divided this by 1000 to get the units into dm^3 which is required to get the amount of moles (mol/dm^3) used to neutralise the acid

Look at step three form here to see it written out clearer.
Titration Calculations (http://www.webchem.net/notes/Periodicity/titration_calcs.htm)

What you have done is taken the volume and multiplied it by the MW