Two spheres of radius a and b are charged and joined by a wire. The ratio of electric fields due to spheres of radius a and b is

Joined by a wire means they are at the same potential.

For same potential $\frac{{\mathrm{kQ}}_1}{{\mathrm{a}}_1}=\frac{{\mathrm{kQ}}_2}{{\mathrm{a}}_2}\Rightarrow \frac{{\mathrm{Q}}_1}{{\mathrm{Q}}_2}=\frac{\mathrm{a}}{\mathrm{b}}\mathrm{ . }$ Further, the electric field at the surface of the sphere having radius R and charge Q is $\frac{\mathrm{kQ}}{{\mathrm{R}}^2}.$

$\therefore \frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}=\frac{{\mathrm{kQ}}_1/{\mathrm{a}}^2}{{\mathrm{kQ}}_2/{\mathrm{b}}_2}=\frac{{\mathrm{Q}}_1}{{\mathrm{Q}}_2}\times \frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=\frac{\mathrm{b}}{\mathrm{a}}$