Let two identical spheres have charge q. And distance between them = r

$$\therefore$$ Force between the spheres $$(F) = {{k{q^2}} \over {{r^2}}}$$

Now when an identical uncharged sphere C comes in contact with A, charge q on sphere A get's divided equally to both sphere.

So, both sphere A and C have charge $$ = {q \over 2}$$

Now, C get's in contact with B. So their total charge $$\left( {q + {q \over 2}} \right)$$ gets divided equally.

So, charge on both B and C is $$ = {{q + {q \over 2}} \over 2} = {{3q} \over 4}$$

Now, C is place midpoint between A and B.

Repulsion force between A and C,

$${F_{AC}} = {{k\left( {{q \over 2}} \right)\left( {{{3q} \over 4}} \right)} \over {{{\left( {{r \over 2}} \right)}^2}}} = {{4k \times 3{q^2}} \over {8{r^2}}}$$

Repulsion force between B and C,

$${F_{BC}} = {{k\left( {{{3q} \over 4}} \right)\left( {{{3q} \over 4}} \right)} \over {{{\left( {{r \over 2}} \right)}^2}}} = {{4k \times 9{q^2}} \over {16{r^2}}}$$

$$\therefore$$ Net force on C,

$${F_{net}} = {F_{BC}} - {F_{AC}}$$

$$ = {{4k \times 9{q^2}} \over {16{r^2}}} - {{4k \times 3{q^2}} \over {8{r^2}}}$$

$$ = {{k{q^2}} \over {{r^2}}}\left( {{9 \over 4} - {3 \over 2}} \right)$$

$$ = {{k{q^2}} \over {{r^2}}} \times {3 \over 4}$$

$$ = {3 \over 4}F$$ [as $$F = {{k{q^2}} \over {{r^2}}}$$]