$\frac{{{E}_{2}}}{{{E}_{1}}}=\left| \frac{{{q}_{2}}}{{{q}_{1}}} \right|\times {{\left( \frac{{{r}_{1}}}{{{r}_{2}}} \right)}^{2}}\to \frac{{{E}_{2}}}{E}=\frac{3q}{q}\times {{\left( \frac{r}{2r} \right)}^{2}}\to {{E}_{2}}=\frac{3}{4}E$