The diameter of the conducting sphere is d = 2 cm.
The radius of the conducting sphere will be
[tex]\begin{gathered} r=\frac{d}{2}=\frac{2\times10^{-2}\text{ m}}{2} \\ =1\text{ }\times10^{-2}\text{ m} \end{gathered}[/tex]The charge on the sphere is
[tex]\begin{gathered} q=\text{ 4 nC } \\ =\text{ 4}\times10^{-9}\text{ C} \end{gathered}[/tex]The force on the surface due to the charge is
[tex]\begin{gathered} F=\frac{1}{4\pi\epsilon_0}\times\frac{q}{r^2} \\ =\frac{9\times10^9\times4\times10^{-9}}{1\times10^{-4}} \\ =36\times10^4\text{ C} \end{gathered}[/tex]So, the surface electric field strength will be
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