Given that $\varepsilon=10 \sin \left(250 \pi \mathrm{s}^{-}\right) \mathrm{t}$
Compair with $\varepsilon=\varepsilon_0 \sin (\omega \mathrm{t})$ $\omega=250 \pi \mathrm{Rad} / \mathrm{s}$
The energy dissipated as
$$
\begin{aligned}
& \mathrm{H}=\frac{1}{\mathrm{R}} \int_0^{10^{-3}} \varepsilon_0^2 \sin ^2 \omega \mathrm{tdt} \\
& =\frac{\varepsilon_0^2}{2 \mathrm{R}} \int_0^{10^{-3}}(1-\cos 2 \omega \mathrm{t}) \mathrm{dt} \\
& =\frac{\varepsilon_0^2}{2 \mathrm{R}}\left[\mathrm{t}-\frac{\sin 2 \omega \mathrm{t}}{2 \omega}\right]^{10^{-3}} \\
& =\frac{\varepsilon_0^2}{2 \mathrm{R}}\left[10^{-3}-\frac{\sin 2 \times 250 \pi}{2 \times 250 \pi} \times 10^{-3}\right] \\
& =\frac{100}{2 \times 100}\left[10^{-3}-\frac{1}{500 \pi}\right] \\
& =\frac{1}{2}\left[\frac{1}{1000}-\frac{1}{500 \pi}\right]=\frac{1}{2}\left[\frac{\pi-2}{1000 \pi}\right] \\
& =\frac{0.57}{\pi} \mathrm{mJ}
\end{aligned}
$$