b
As the capacitors are connected in parallel, therefore potential difference across both the condensors remains the same.
$\therefore Q_{1}=C V ;$
$Q_{2} =\frac{C}{2} V $
$\text { Also }, Q =Q_{1}+Q_{2}$
$ = CV+\frac{C}{2} V=\frac{3}{2} C V$
Work done in charging fully both the condensors is given by
$W=\frac{1}{2} Q V=\frac{1}{2} \times\left(\frac{3}{2}\, C V\right) V=\frac{3}{4} \,C V^{2}$