= Solution
For forward and reverse elementary reactions at equilibrium, the two rates are equal. Both directions have two reactant particles and use the same <concentration> convention, so the conversion factors cancel and $K_{\rm eq}=k_3/k_{-3}$. Therefore
$$
\boxed{k_{-3}=\frac{1.933\times10^{-14}}{7.244\times10^{33}}
=2.668\times10^{-48}\,\mathrm{dm^3\,molecule^{-1}\,s^{-1}}.}
$$
For molar rather than molecular number <concentration>, the bimolecular <rate constant> is multiplied by the <Avogadro constant>:
$$
\boxed{k_{-3}=2.668\times10^{-48}(6.022\times10^{23})
=1.607\times10^{-24}\,\mathrm{dm^3\,mol^{-1}\,s^{-1}}.}
$$
Only one factor is needed: converting the two <concentration> factors contributes $N_A^2$, while converting the production rate from <molecules> to moles removes one factor.
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