= Solution
The <Parseval identity> for this <Fourier transform> convention is
$$
\boxed{\int_{\mathbb R}|h(x)|^2dx=\frac1{2\pi}\int_{\mathbb R}|\widetilde h(k)|^2dk.}
$$
Apply it to $h=g$. Its squared <norm> is $2\int_0^1(1-x)^2dx=2/3$, so
$$
\frac23=\frac1{2\pi}\int_{\mathbb R}\frac{16\sin^4(k/2)}{k^4}\,dk.
$$
Under $k=2x$, the last integrand times $dk$ becomes $2\sin^4x\,dx/x^4$. Consequently
$$
\boxed{\int_{-\infty}^{\infty}\frac{\sin^4x}{x^4}dx=\frac{2\pi}{3}.}
$$
The integrand has continuous value $1$ at zero and decays integrably at infinity, so this is an ordinary convergent <improper integral>.
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