In the given circuit diagram, on closing the switch S 

Initial effective capacitance, $C_1=\frac{C}{2}$
Final effective capacitance, $C_2=\frac{2C\times C}{2C+C}=\frac{2C}{3}$
Charge passed through the battery on closing the switch S,  $\Delta q=C_{2}V-C_{1}V=\left(\frac{2C}{3}-\frac{C}{2}\right)V=\frac{CV}{6}$
Work done by the battery on closing the switch S, $W=V\Delta q=\frac{C{V}^2}{6}$
             Option (c) is correct. Option (a) is wrong.

$W=\text{Heat}+U_2-U_1$

$\text{Heat}=W-U_2+U_1=\frac{C{V}^2}{6}-\frac{1}{2}{C}_2{V}^2+\frac{1}{2}{C}_1{V}^2$

$=\frac{C{V}^2}{6}-\frac{1}{2}\frac{2C}{3}{V}^2+\frac{1}{2}\frac{C}{2}{V}^2$

$=\frac{C{V}^2}{6}-\frac{C{V}^2}{3}+\frac{C{V}^2}{4}$

$=C{V}^2\left(\frac{2-4+3}{12}\right)=\frac{C{V}^2}{12}$       Option (d) is correct. Option (b) is wrong.