I need just a hint please. It seems that I have to prove that $f(x)=mx$ in which $m\in \mathbb{R}.$ But I couldn't handle it.

Problem: We say that a sequence $x_{n}\; , n = 1, 2,\cdots ,$ Cesaro converges to $a,$ if $$\lim\limits_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n}x_{i}=a.$$ A function $f$ is Cesaro continuous at $a$ if $x_{n}\to a$ (in Cesaro mean) implies $f(x_{n})\to a$ (in Cesaro mean). Prove that, if $f : \mathbb{R}\to\mathbb{R}$ is Cesaro continuous at $0$ and $f(0) = 0,$ then $f$ is linear.

I guess you mean to say that $f(x_n)\to f(a)$ whenever $x_n\to a$, all in the Cesaro sense.
If this is the case, let $x\in\mathbb R$ and consider the sequence $x_n=(-1)^nx$. Then $x_n\to 0$ in the Cesaro sense, therefore $$f(x)+f(-x)=\lim_{n\to\infty}\frac{1}{2n}\sum_{k=1}^{2n}f(x_k)=f(0)=0,$$ therefore $f$ is odd.
Note now that, if $x,y\in\mathbb R$ and $(y_n)$ is the sequence $$x+y, -x,-y, x+y, -x, -y,\dots,$$ then $y_n\to 0$ in the Cesaro sense, therefore, as above, $$f(x+y)+f(-x)+f(-y)=\lim_{n\to\infty}\frac{1}{3n}\sum_{k=1}^{3n}f(y_k)=f(0)=0,$$ therefore $$f(x+y)=-f(-x)-f(-y)=f(x)+f(y),$$ since $f$ is odd. This shows that $f(x)=ax$ for some $a$, at least when $x\in\mathbb Q$.
Finally, for $r$ being irrational, let $(x_n)$ be a sequence of rational numbers that converges to $r$, and consider the sequence $(z_n)$, which is defined by $$x_1,-r,x_2, -r,x_3,-r,\dots .$$ Then $z_n\to 0$ in the Cesaro sense, and as above you can show that $f(r)=ar$.