= Solution
For $m\geq1$, let $S_m$ be the set of vertices of $B(m)$ connected to $\partial B(m)$ by an open <path in a graph> using only edges of $B(m)$. Every infinite-cluster vertex of $B(m)$ belongs to $S_m$, by stopping an infinite open <path in a graph> at its first boundary visit. Hence the <increasing event>
$$
E_m=\{|S_m|\geq\theta(p)(2m+1)^2/2\}
$$
has probability at least $\theta(p)/2$ by part (a). Importantly, $E_m$ depends only on the edges with both endpoints in $B(m)$.
Open all $8m$ perimeter edges joining successive vertices of $\partial B(m)$. They form a <cycle in a graph>, and opening them joins every vertex of $S_m$ into one <percolation cluster>, containing the specified perimeter vertex $v_m=(m,0)$. The <Harris-FKG inequality> bounds the probability of this event together with $E_m$ below by $p^{8m}\theta(p)/2$.
Close all $8m+4$ edges in the <edge boundary> of $B(m)$. These edges are distinct from the internal edges already considered, so their states are independent of those events. The resulting <percolation cluster> of $v_m$ is finite, contained in $B(m)$, and has at least $|S_m|$ vertices. Therefore, by translating $v_m$ to the origin,
$$
\mathbb P_p\bigl(\theta(p)(2m+1)^2/2\leq|C|<\infty\bigr)\geq\frac{\theta(p)}2p^{8m}(1-p)^{8m+4}.
$$
For $n\geq1$, choose $m=\lceil\sqrt{n/(2\theta(p))}\rceil$. Then the threshold is at least $n$, and $m\leq(1+(2\theta(p))^{-1/2})\sqrt n$. Put $A=-\log p$ and $B=-\log(1-p)$; both are positive. The preceding lower bound is at least $e^{-c_1\sqrt n}$ with, for example,
$$
\boxed{c_1=8(A+B)\left(1+\frac1{\sqrt{2\theta(p)}}\right)+4B+\log\frac2{\theta(p)},\qquad
\mathbb P_p(n\leq|C|<\infty)\geq e^{-c_1\sqrt n}.}
$$
The construction modifies only order-$m$ boundary edges while trapping order-$m^2$ vertices, which explains the <stretched exponential> scale. Here $\mathbb N$ is interpreted as the positive integers; the proposed lower bound at $n=0$ would be false because $\theta(p)>0$.
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