Let a be the radius of a sphere $A, Q_{A}$ be the charge on the sphere, and $C_{A}$ be the capacitance of the sphere.

Let $b$ be the radius of a sphere $B$, $Q$ be the charge on the sphere, and $C$ a be the capacitance of the sphere.

since the two spheres are connected with a wire, their potential $(v)$ will become equal.

Let Eabe the electric field of sphere $A$ and $E_{8}$ be the electric field of sphere $B$. Therefore, their ratio,

$\frac{E_{A}}{E_{B}}=\frac{Q_{A}}{4 \pi \epsilon_{0} \times a_{2}} \times \frac{b^{2} \times 4 \pi \epsilon_{0}}{Q_{B}}$

$\frac{E_{A}}{E_{B}}=\frac{Q_{1}}{Q_{B}} \times \frac{b^{2}}{a^{2}}\ldots(i)$

However, $\frac{Q_{A}}{Q_{B}}=\frac{C_{A} V}{C_{B} V}$

And, $\frac{C_{A}}{C_{B}}=\frac{a}{b}$

$\therefore \frac{Q_{A}}{Q_{B}}=\frac{a}{b}\dots (ii)$

Putting the value of $(ii)$ in $(i)$, we obtain

$\therefore \frac{E_{A}}{E_{B}}=\frac{a}{b} \frac{b^{2}}{a^{2}}=\frac{b}{a}$

Therefore, the ratio of electric fields at the surface is $b/a.$