Distance between the spheres, $A$ and $B, r=0.5\,m$

Initially, the charge on each sphere, $q=6.5 \times 10^{-7} \,C$

When sphere $A$ is touched with an uncharged sphere $C, q / 2$ amount of charge from $A$ will transfer to sphere $C$. Hence, charge on each of the spheres, $A$ and $C ,$ is $q / 2 .$

When sphere $C$ with charge $q / 2$ is brought in contact with sphere $B$ with charge $q$, total charges on the system will divide into two equal halves given as,

$\frac{1}{2}\left(q+\frac{q}{2}\right)=\frac{3 q}{4}$

Hence, charge on each of the spheres, $C$ and $B ,$ is $\frac{3 q}{4}$ Force of repulsion between sphere A having charge $q / 2$ and sphere $B$ having charge $\frac{3 q}{4}$ is $F=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{A} q_{B}}{r^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\frac{q}{2} \times \frac{3 q}{4}}{r^{2}}$

$=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{3 q^{2}}{8 r^{2}}=\frac{9 \times 10^{9} \times 3 \times\left(6.5 \times 10^{-7}\right)^{2}}{8 \times(0.5)^{2}}$$=5.703 \times 10^{-3}\, N$

Therefore, the force of attraction between the two spheres is $5.703 \times 10^{-3} \;N$.