Given that, line makes angles $\alpha, \beta, \gamma, \delta$ with the four diagonals of a cube, then we know that
$\begin{aligned}
& \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma+\cos ^2 \delta=\frac{4}{3} \\
& \text { Since, } \cos ^2 \theta=1-\sin ^2 \theta \\
& 1-\sin ^2 \alpha+1-\sin ^2 \beta+1-\sin ^2 \gamma+1-\sin ^2 \delta=\frac{4}{3} \\
& \therefore \quad \sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma+\sin ^2 \delta=4-\frac{4}{3}=\frac{8}{3}
\end{aligned}$