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 $B, Q_B$ be the charge on the sphere, and $C_B$ be the capacitance of the sphere. Since the two spheres are connected with a wire, their potential (V) will become equal.
Let $E_A$ be the electric field of sphere A and $E_B4 be the electric field of sphere B. Therefore, their ratio,
$\large\frac{E_A}{E_B}= \large\frac{Q_4}{4 \pi \in_0 \times a_2} \times \frac{b^2 \times 4 \pi \in_0}{Q_8}$
$\large\frac{E_A}{E_B} =\frac{Q_A}{Q_B} \times \frac{b^2}{a^2}$ ------------(1)
How ever $\large\frac{Q_A}{Q_B} = \frac{C_1 V}{C_B V}$
and $\large\frac{C_A}{C_B} = \frac{a}{b}$----------------(2)
Putting the value of (2) in (1), we obtain
$\large\frac{E_A}{E_B} =\large\frac{a}{b} \frac{b^2}{a^2}=\frac{a}{b}$
Therefore, the ratio of electric fields at the surface is $\large\frac{b}{a}$