We must assume that $|A|>1$ which implies that $R(a) \le 2$ for all $a \in A$.

Let $V_d$ denote the volume of he unit ball $B_1=B(0,1)$. The open balls $\{B(a,R(a)/2)\}_{a \in A}$ are pairwise disjoint and contained in $B(0,2)$. Comparing the volume of their union to the volume of $B(0,2)$, we infer that $2^d V_d$. $$\sum_{a \in A} R(a)^d \, 2^{-d}\, V_d\, \le \, 2^d \,V_d\,.$$ Multiplying both sides by $2^d$, dividing by $|A| V_d$, and switching the two sides, we obtain that $$|A|^{-1}4^d \ge |A|^{-1}\sum_{a \in A} R(a)^d \ge \Bigl(|A|^{-1}\sum_{a \in A} R(a)\Bigr)^d = \ell ^d \,,$$ where we used convexity of $x \mapsto x^d$ in the second inequality. This proves the inequality claimed.

We must assume that $|A|>1$ which implies that $R(a) \le 2$ for all $a \in A$.

Let $V_d$ denote the volume of he unit ball $B_1=B(0,1)$. The open balls $\{B(a,R(a)/2)\}_{a \in A}$ are pairwise disjoint and contained in $B(0,2)$. Comparing the volume of their union to the volume of $B(0,2)$, we infer that 
$$ 2^dV_d \ge \sum_{a \in A} R(a)^d \, 2^{-d}\, V_d\,.$$ Multiplying both sides by $2^d$ and dividing by $|A| V_d$, we obtain that $$|A|^{-1}4^d \ge |A|^{-1}\sum_{a \in A} R(a)^d \ge \Bigl(|A|^{-1}\sum_{a \in A} R(a)\Bigr)^d = \ell ^d \,,$$ where we used convexity of $x \mapsto x^d$ in the second inequality. This proves the claim $|A| \le (4/\ell)^d$.