the following problem says: Show that if is f an integrable function in $\mathbb{R}^d$ and not identically null, then $$f^*(x)\geq\frac{c}{|x|^d}$$ where $c>0$, $|x|\geq 1$ and $f^*(x)=\sup_{x\in B}\frac{1}{m(B)}\int_B|f(y)|dy$. Conclude that $f^*$ is not integrable.
I've found the bound $c=\frac{1}{m(B)}|\int_B f(y)dy|$, it's ok?
You did not say what $B$ is. As long as $f$ is not identically zero there exists some $R > 0$ with the property that $$\int_{B(0,R)} |f(y)| \, dy > 0.$$ If $x \in \mathbb R^d$ then $B(0,R) \subset B(x,|x|+R)$ so that $$ \int_{B(x,|x|+R)} |f(y)| \, dy \ge \int_{B(0,R)} |f(y)| \, dy $$ and in particular $$ f^*(x) \ge \frac1{m(B(x,|x| + R))} \int_{B(0,R)} |f(y)| \, dy = \frac1{\alpha(d)(|x|+R)^d}\int_{B(0,R)} |f(y)| \, dy.$$ Since $|x|/(|x| + R) \ge 1/(1+R)$ whenever $R > 0$ and $|x| \ge 1$ you get $$ f^*(x) \ge \frac{1}{\alpha(d) (1+R)^d |x|^d} \int_{B(0,R)} |f(y)| \, dy,\ |x| \ge 1.$$