The reaction that occurs can be represented as
$ CuCO_3+H_2SO_4 \to CuSO_4+H_2O+CO_2 \uparrow $
From the above equation we find that $123.5 \;g$ of $CuCO_3$ requires $H_2SO_4=98\;g$.
$0.5 \;g \;CucO_3$ will require $H_2SO_4 =\large\frac{98}{123.5} $$ \times 0.5 =0.39676\;g$
To find the volume of $H_2SO_4$ solution required use the reaction.
$Molarity =\large\frac{weight \;of\;solute (H_2SO_4)}{mol.weigth\;of\;solute (H_2SO_4)} \times \frac{1000}{Volume \;of\;solution}$
$0.5 =\large\frac{0.39676}{98} \times \frac{1000}{V}$
or $V= \large\frac{0.396 \times 6 \times 1000}{98 \times 0.5}$
$\qquad= 8.097\;ml$
Hence c is the correct answer.