c
$V =\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{r}=Q \cdot 10^{11} \text { volts }$
$\therefore \frac{1}{r} =4 \pi \varepsilon_{0} \cdot 10^{11}$
$E =\frac{\text { potential }}{r}=Q \cdot 10^{11} \times 4 \pi \varepsilon_{0} \cdot 10^{11}$
$\Rightarrow E=4 \pi \varepsilon_{0} \cdot Q \cdot 10^{22} \mathrm\,{volt/m}$