Solution :
Electric field E just outside the conductor is given by the relation,
$E= \large\frac{1}{4 \pi \in_0}.\frac{q}{r^2}$
Where , $\in_0$=Permittivity of free space and $\large\frac{1}{4 \pi \in_0}$$ = 9 \times 10^9 Nm^2C ^{-2}$ Therefore,
$E= \large\frac{9 \times 10^9 \times 1.6 \times 10^{-7}}{(0.12)^2} $$=10^5\;NC^{-1}$
Therefore, the electric field just outside the sphere is $10^5 NC^{ -1 }$.