In iodometry , $K_2Cr_2O_7$ acts as an oxidising agent to liberate iodine $(I_2)$ from an iodide such as $KI$ or $NaI$
The iodine $(I_2) $ thus liberated. is titrated with hypo $(Na_2S_2O_3)$ solution.
$Na_2S_2O_3+I_2 \longrightarrow Na_2S_4O_6 +2 NaI$
We find that one mole of $K_2Cr_2O_7$ accepts 6 moles of electron as follows.
$Cr_2O_7^{2-}+14H^{+}+6e^{-} \longrightarrow 2Cr{3+}+7H_2O$
Thus equivalent weight $=\large\frac{Molecular \;weight}{6}$
Hence b is the correct answer.