Answer
$\Delta N = 2.5 \times 10^{11}$
Work Step by Step
The number of decay during the 12 hour is
$\Delta N = N_o (1 - e^{-tln2/T_{1/2}})$
Where
$t = 12 h$
$T_{1/2} = 24100 y \times 8760 h/y = 2.11 \times 10^8 h$
*Note that we use hour as time unit in the equation
So
$\Delta N = (6.3 \times 10^{18}) (1 - e^{-(12 h)ln2/2.11 \times 10^8h})$
$\Delta N = 2.5 \times 10^{11}$
