Let $S$ be subset of $\mathbb{R}^n$ with perimeter 1.

Isoperimetric inequality states that then the volume of $S$ is not greater than $V_n$,

where $V_n$ is the volume of a ball in $\mathbb{R}^n$ with perimeter 1.

Assume that $C \cdot \text{[Volume of }S] \ge V_n$, where $C$ is some constant.

Is it true that 99% of $S$ can be covered by union of constant number of balls with constant radius? (The constants can be depend on $n$ and $C$.)

P.S. The question is motivated by the similar question about boolean cube.

UPD: I can prove this statement for $n=2$: if $F$ belongs to $\mathbb{R}^2$, the perimeter of $F$ is equal to $1$ and its area is at least $\frac{1}{C}$ then 99% of $F$ can be covered by $1000C$ balls with radius $1$.

Indeed, consider $D:=100C$ connectivity components of $F$ with the greatest perimeter. Of course, they can be coveres by $100C$ balls with radius $1$. Our goal is to show that other connectivity components of $F$ cannot have total area $\frac{1}{100C}$ or bigger. Denote the set of non-covered components by $\mathcal{H}$. We can assume that all elements of $\mathcal{H}$ are balls---this is the worst case scenario. More over we can assume that all---or all except one---balls have the same (the greatest) perimeter (because $(r+a)^2 + (l-a)^2 > r^2 + l^2$ if $r > l$ and $a>0$). So, if $|\mathcal{H}|=m$ then we have $m$ balls with perimeter at most $\frac{1}{D+m}$, the total area of these balls is less than $\frac{1}{100C}$ since $D = 100C$.