Peak and RMS voltage for a sine wave
I know Vpk=sqrt(2)Vrms


That is only valid for certain waveforms. It is well-known to be valid for a sinewave, it is not so well known to be true for the waveform of part B in your question.

BTW you seem to have read the question backwards, it's asking what value of \$V_{peak}\$ is needed to make \$V_{rms} = 140 V\$ NOT what the value of \$V_{rms}\$ is for \$V_{peak}=140V\$


  My question is: How do we find the area under the graph?. We square the wave as before, obtaining the negative cycle as a squared positive one. 


For simplicity let \$t = \frac{1}{8}T\$ . The equation for the instantanious voltage of the sinewave is then.

$$V=V_{sin}\sin(2\pi t)$$

We calculate the RMS by first squaring, then calculating the mean over one cycle (which since it's a continuous function we do by integration) and then finally square rooting.

$$V_{RMS}=\sqrt{\int_0^1(V_{sin}\sin(2\pi t))^2\mathrm{d}t}$$

Then it's a matter of trig rearrangements and integration to solve that.

But honestly I would expect your examiner to expect you to just know that \$V_{RMS} = \frac{V_{sin}}{\sqrt{2}}\$ and not have to re-derive it in the exam.


  I see we have 2 flipped parabolas.


Not really, what you find when you do the trig identies is that squaring a sinewave ends up with a sinewave plus a constant.