How many clauses are required for SAT to be NP-hard in CNF formulas?
Lower bound. For $g \le c \cdot \sqrt{\log n}$ there exists a polynomial-time algorithm. The idea is the following: if some clauses have too many variables, then it should be trivial to select some variable to satisfy this clause, without hurting clauses with few variables. We repeat the following:
Find the clause with the smallest number of variables. Let $x_1,\ldots,x_k$ be the variables participating in this clause.

If $k &gt; g$, then the entire formula is satisfiable (we process clauses one by one and select a variable which we didn't select before).
Otherwise, we remove the clause. We also remove $x_1,\ldots,x_k$ from all other clauses.

Now, we have to satisfy the removed clauses. Since there are at most $g$ clauses and each of them introduces at most $g$ new variables, it means that there are at most $g^2 = c^2 \cdot \log n$ variables overall. Therefore, there are at most $n^{c^2}$ variable combinations, and we can just use brute-force.
Conditional upper bound. It is almost tight in the following sense. Assume that the lower bound on the SAT with $n$ variables and $\ge c\cdot n$ clauses (for some $c$, e.g. coming from $3$-coloring) is $\alpha^n$ ($\alpha \in (1, 2]$). Note that the same lower bound holds after our transformation (since we can just apply it before any algorithm). Therefore, if there are at least $\log^{1+\epsilon} n$ clauses, they can have $\frac {\log^{1+\epsilon} n} c$ variables and the lower bound for running time for our problem is
$$\alpha^{\frac {\log^{1+\epsilon} n} c} = n^{\frac {\log^\epsilon n \cdot \log \alpha} {c}},$$
which is super-polynomial.