Correct Answer: 2

Solution : Given: $a^{2}+b^{2}+c^{2}=ab+bc+ca$
⇒ $a^{2}+b^{2}+c^{2}-ab-bc-ca=0$
⇒ $\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]=0$
⇒ $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$
The square of any real number is always non-negative.
Therefore, we have:
$(a-b)^{2}=0 \Rightarrow a=b$
$(b-c)^{2}=0 \Rightarrow b=c$
$(c-a)^{2}=0 \Rightarrow c=a$
Thus, we have $a=b=c$. Now, we can calculate $\frac{a+c}{b}$ as follows:
$\frac{a+c}{b}=\frac{a+a}{a}=2$
Hence, the correct answer is 2.