Welcome to MO! However, your post needs to be edited with care. First, the $X_n$'s are iid only when $\sigma_n$ does not depend on $n$. Second, you should not write $X_n\in N(0,\sigma_n)$, because $s\in S$ can only mean that $s$ is an element of a set $S$. Third, the standard notation for the normal distribution is $N(\mu,\sigma^2)$, rather than $N(\mu,\sigma)$. Fourth, the expression $P[\delta>\frac{\{k\le n: X_k\le c_k\}}{b_n}>\epsilon]$ does not make sense, since $\{k\le n: X_k\le c_k\}$ is a random set, rather than a real-valued random variable (r.v.). So, apparently your question is about a condition for

$$P_n:=P\Big(\delta>\frac{S_n}{b_n}>\epsilon\Big)\to 1$$
as $n\to\infty$, where $S_n=\sum_{k=1}^n Y_k=\sum_{k=1}^n 1_{X_k\le c_k}=\#\{k\le n: X_k\le c_k\}$ (as in your post), $Y_k:=1_{X_k\le c_k}$, and the $X_k$'s are independent normal zero-mean r.v.'s with variances $\sigma_k^2$. Let $p_{k}:=EY_k=P(X_k\le c_k)=\Phi(c_k/\sigma_k)$, where $\Phi$ is the standard normal cdf, so that $ES_n=\mu_n:=\sum_1^n p_k$ and $\sqrt{Var\,S_n}=B_n:=\sqrt{\sum_1^n (1-p_k)p_k}$. Then, letting
$$b_n:=s\mu_n,\quad s:=\frac2{\epsilon+\delta}, \tag{1}
$$
assuming
$$t:=\frac{\delta-\epsilon}2>0,
$$
and using Chebyshev's inequality, we have
$$P_n=P\Big(\delta-\frac{\mu_n}{b_n}>\frac{S_n-\mu_n}{b_n}>\epsilon-\frac{\mu_n}{b_n}\Big)
=P\Big(\Big|\frac{S_n-\mu_n}{s\mu_n}\Big|<t\Big)\ge1-\frac{B_n^2}{\mu_n^2s^2t^2}.
$$
So, we will have $P_n\to1$ if $b_n$ is as in (1) and

$$\frac{B_n^2}{\mu_n^2}=\frac{\sum_1^n (1-p_k)p_k}{\big(\sum_1^n p_k\big)^2}\to0,
$$
which will be the case if e.g. the $p_k$'s are bounded away from $0$ or, more generally, if $\sum_1^n p_k\to\infty$ as $n\to\infty$ (because $\sum_1^n (1-p_k)p_k\le\sum_1^n p_k$).