Potential of a charged sphere is given by the formula
$V =\frac{ kQ }{ R }$
Hence, by $Q = CV$,
$C =\frac{ Q }{ V }=\frac{ R }{ k }$
As per given information, $\frac{ R }{ k }=1 \mu F$
After splitting,
$8\left(\frac{4}{3} \pi r ^{3}\right)=\frac{4}{3} \pi R ^{3}$
$r =\frac{ R }{2}$
Thus, capacitance of each small drop is $V =\frac{ r }{ k }=\frac{1}{2} \frac{ R }{ k }=\frac{1}{2} \mu F$