Check the answer by building a known hadron from d-quarks. The neutron is the composite $udd$, and its total charge must be zero.
Using the up-quark charge $+\tfrac{2}{3}e$, the neutron charge is $\left(+\tfrac{2}{3}\right)+\left(-\tfrac{1}{3}\right)+\left(-\tfrac{1}{3}\right)=0$, which is consistent only if each d-quark carries $-\tfrac{1}{3}e$. Any other value listed (such as $+\tfrac{2}{3}e$ or $+\tfrac{1}{3}e$) would give a nonzero neutron charge, so those options are ruled out.
For strangeness: this quantum number counts the number of strange quarks (with a minus sign), so $S=-(n_{s}-n_{\bar s})$. A d-quark contains no strange content, hence $S=0$. Only the s-quark and its antiquark change strangeness. Both properties together match option (C).
\[\boxed{Q_{d}=-\tfrac{1}{3}e,\quad S_{d}=0}\]