The answer is no. Let $m$ be a very large positive constant. You can find smooth $f$ that equals $m$ on a small ball $B_R(0)$ and still satisfy $\int_{B_6(0)}|f|<\delta$. You can do it with $R$ such that $R^n= \delta 2^{-1}m^{-1}\omega_n^{-1}$, where $\omega_n$ is the volume of the unit ball. Indeed, you approximate the $m\chi_{B_R(0)}$ by a smooth function.

Then $$ Mf(0)\geq R^{4-n}\int_{B_R(0)}f=C(n)\delta^{\frac{4}{n}}m^{1-\frac{4}{n}}. $$ Note that $m$ is independent of $\delta$ and it can be arbitrrily large, meaning that $Mf(0)$ can be arbitrarily large and so $Mf$ will be arbitrarily large in a neighborhood of $0$. That also means that a neighborhood of $0$ will be in the set $A$.

The answer is no. Let $m$ be a very large positive constant. You can find smooth $f$ that equals $m$ on a small ball $B_R(0)$ and still satisfy $\int_{B_6(0)}|f|<\delta$. You can do it with $R$ such that $R^n= \delta 2^{-1}m^{-1}\omega_n^{-1}$, where $\omega_n$ is the volume of the unit ball. Indeed, you approximate the $m\chi_{B_R(0)}$ by a smooth function.

Then $$ Mf(0)\geq R^{4-n}\int_{B_R(0)}f=C(n)\delta^{\frac{4}{n}}m^{1-\frac{4}{n}}. $$ Note that $m$ is independent of $\delta$ and it can be arbitrrily large, meaning that $Mf(0)$ can be arbitrarily large and so $Mf$ will be arbitrarily large in a neighborhood of $0$. That also means that a neighborhood of $0$ will be in the set $A$. Hence $G$ will not be dense and $\mathcal{H}^n(A)>0$.