Two charged conducting spheres of radii a and b are connected by a wire. The ratio of the electric fields at their surfaces respectively, is

$b:a$

The correct answer is Option (3) → $b:a$

Let the charges on the spheres be $Q_1$ and $Q_2$, and radii be $a$ and $b$.

When two conducting spheres are connected by a wire, they attain the same potential:

$V_1 = V_2$

$\frac{Q_1}{4 \pi \epsilon_0 a} = \frac{Q_2}{4 \pi \epsilon_0 b}$

$\frac{Q_1}{a} = \frac{Q_2}{b}$

$Q_1 : Q_2 = a : b$

The electric field at the surface of a sphere is:

$E = \frac{Q}{4 \pi \epsilon_0 r^2}$

Ratio of electric fields at the surfaces:

$\frac{E_1}{E_2} = \frac{Q_1 / a^2}{Q_2 / b^2} = \frac{a/b}{a^2/b^2} = \frac{b}{a}$