d
In air \(\tan \frac{\theta}{2}=\frac{F}{m g}=\frac{q^2}{4 \pi \varepsilon_0 \mathrm{r}^2 m g}\)

In water \(\tan \frac{\theta}{2}=\frac{\mathrm{F}^{\prime}}{\mathrm{mg}^{\prime}}=\frac{\mathrm{q}^2}{4 \pi \varepsilon_0 \varepsilon_{\mathrm{r}} \mathrm{r}^2 \mathrm{mg}_{\text {eff }}}\)

Equate both equations

\(\varepsilon_0 g=\varepsilon_0 \varepsilon_{\mathrm{r}} \mathrm{g}\left[1-\frac{1}{1.5}\right]\)

\(\varepsilon_{\mathrm{T}}=3\)