Solution

We assume the spheres are far apart. Then the charge distribution on each of them is spherically symmetric and Coulomb's law can be used. Let $$q_1$$ and $$q_2$$ be the original charges. We choose the coordinate system so the force on $$q_2$$ is positive if it repelled by $$q_1$$. Then, the force on $$q_2$$ is
$$F_a=-\dfrac {1}{4\pi { \varepsilon }_{ 0 }}\dfrac {q_1 q_2}{r^2}=-k\dfrac {q_1 q_2}{r^2}$$
where $$r=0.0500\ m$$. The negative sign indicates that the spheres attract each other. After the wire is connected, the spheres, being identical, acquire the same charge. Since charge is conserved, the total charge is the same it was originally. This means the charge on each sphere is $$(q_1 +q_2)/2$$. The force is now one of repulsion and is given by
$$F_b =\dfrac {1}{4\pi { \vare
psilon }_{ 0 }}\dfrac {\left(\dfrac {q_1+q_2}{2}\right)\left(\dfrac {q_1 +q_2}{2}\right)}{r^2}=k\dfrac {(q_1 +q_2)^2}{4r^2}$$
We solve the two force equations simultaneously for $$q_1$$ and $$q_2$$. The first gives the product
$$F_b =\dfrac {1}{4\pi { \varepsilon }_{ 0 }}\dfrac {\left(\dfrac {q_1+q_2}{2}\right)\left(\dfrac {q_1 +q_2}{2}\right)}{r^2}=k\dfrac {(q_1 +q_2)^2}{4r^2}$$
We solve the two force equations simultaneously for $$q_1$$ and $$q_2$$. The first gives the product
$$q_1q_2=-\dfrac {r^2 F_q}{k}=-\dfrac {(0.500\ m)^2 (0.108\ N)}{8.99\times 10^{9}N.m^2 /C^2}=-3.00\times 10^{-12} C^2$$
and the second gives the sum
$$q_1+q_2=2r\sqrt {\dfrac {F_b}{k}}=2(0.500\ m)\sqrt {\dfrac {0.360\ N}{8.99\times 10^9 N. m^2 /C^2}}=2.00\times 10^{-6}C$$
where we have taken the positive root. Thus, the product results provides the relation
$$q_2=\dfrac {-(3.00\times 10^{-12}C^2)}{q_1} $$
which we substitute into the sum result, producing
$$q_1-\dfrac {3.00\times 10^{-12} C^2}{q_1}=2.00\times 10^{-6}C$$.
Multiplication by $$q_1$$ and rearranging, we obtain a quadratic equation
$$q_1^2 -(2.00\times 10^{-6}C)q_1-3.00\times 10^{-12}C^2=0$$
The solutions are
$$q_1=\dfrac {2.00\times 10^{-6}C\pm \sqrt {(-2.00\times 10^{-6}C)^2}-4(3.00\times 10^{-12}C^2)}{2}$$
If the positive sign is used, $$q_1=3.00\times 10^{-6}C$$, and if the negative is used, $$q_1=-1.00\times 10^{-6}C$$.
If we insted work with the $$q_1=-1.00\times 10^{-6}C$$, then we find $$q_2=3.00\times 10^{-6}C$$.
Note that since spheres are identical, the solutions are essentially the same: one sphere originally had charge $$-1.00\times 10^{-6}C$$ and the other had charge $$+3.00\times 10^{-6}C$$.
What if we had not made the assumption, above, that $$q_1+q_2 \ge 0$$ ? If the signs of the charges were reserved (so $$q_1+q_2 < 0$$), then the forces remain the same, so a charge of $$+1.00\times 10^{-6}C$$ on one sphere and a charge of $$-3.00\times 10^{-6}C$$ on the other also satisfies the condition of the problem.