Could be instructive to do some sums...
If one takes the max ELI for a circuit protected by a B32 MCB as 1.44 ohms (that's the 17th edition figure before it was slightly reduced by the Cmin reductions in amendment 3).
Also lets assume the resistance of a 0.7mm^2 cable is in the order of 26 mOhms/m (1.0mm^2 CSA has a tabulated value of 18.10 mOhm/m). We can add a further 46 mOhms to the circuit total.
That gets us a ELI with a fault at the end of the lead of 1.49 Ohms.
Our prospective fault current will therefore be 230 / 1.49 = 154A
That is slightly shy of the 160A needed for a "instant" trip on the MCB,
so let's look at the fuse:
That looks well below the 0.1 sec time on that. So lets take a pessimistic worst case of 0.1 secs.
So the remaining question is what size conductor is required to carry
154A for 0.1 sec? If we assume PVC insulation hence a k factor of 115 we get:sqrt( 154^2 x 0.1 ) / 115 = 0.42 mm^2
So we can conclude that the flex will withstand that current for long enough to blow the plug fuse.