Sphere A has a radius of 5.0 cm and carries 12 nC…

Step 1. The charges. $\dfrac{q_A}{q_B} = \dfrac{5.0\ \text{cm}}{15\ \text{cm}} = \dfrac{1}{3}$. So $q_A = \dfrac{1}{4} \times 12\ \text{nC} = 3.0\ \text{nC}$ and $q_B = 9.0\ \text{nC}$.

Step 2. The charge that crosses the wire. Sphere B gained all of its $9.0$ nC from A, so $9.0$ nC crosses.

Step 3. The potential. $V = \dfrac{kq_A}{R_A}$, and $kq_A = 9.0 \times 10^{9} \times 3.0 \times 10^{-9} = 27\ \text{N m}^2/\text{C}$. So $V = \dfrac{27\ \text{N m}^2/\text{C}}{0.050\ \text{m}} = 540\ \text{V}$. Before the wire, sphere A alone, with its $12$ nC, was at $\dfrac{kQ}{R} = 2160\ \text{V}$.

Step 4. The charge on each square metre. $\dfrac{\sigma_A}{\sigma_B} = \dfrac{R_B}{R_A} = \dfrac{15\ \text{cm}}{5.0\ \text{cm}} = 3.0$.