In the figure above

$$
\(\begin{aligned}
& \angle \mathrm{ABC}=\angle \mathrm{BAE}=58^{\circ}(\text { Alternate angles }) \& \\
& \angle \mathrm{BAC}=\angle \mathrm{ABC}=58^{\circ}(\because \mathrm{AC}=\mathrm{BC} \Rightarrow \angle \mathrm{BAC}=\angle \mathrm{ABC})
\end{aligned}\)
$$


Now, as the sum of the angles of a triangle is $\(180^{\circ}\)$, in triangle ABC , we get

$$
\(\angle \mathrm{ACB}=180^{\circ}-(\angle \mathrm{ABC}+\angle \mathrm{BAC})=180^{\circ}-\left(58^{\circ}+58^{\circ}\right)=180^{\circ}-116^{\circ}=64^{\circ}\)
$$


Hence column $\(A\)$ has higher quantity when compared with column $\(B\)$, so the answer is $\((A)\)$.