= Solution
The <binomial upper bound for a Ramsey number> gives $R(3,t)\leq\binom{t+1}{2}=O(t^2)$. For the <triangle-free Ramsey lower bound by alteration>, a <binomial random graph> $G(n,p)$ has expected triangle count $\binom n3p^3$ and expected independent $t$-set count $\binom nt(1-p)^{\binom t2}$. Choose a realization whose sum of these counts is at most its expectation. Delete one vertex from each surviving forbidden set, iterating through the originally listed sets. The remaining <graph> has neither a triangle nor an independent $t$-set. This <random alteration method> proves
$$
R(3,t)>n-\binom n3p^3-\binom nt(1-p)^{\binom t2}.
$$
For large $t$, set $n=\lfloor(t/(4\log t))^{3/2}\rfloor$ and $p=n^{-2/3}$. The triangle term is at most $n/6$. Using $\binom nt\leq(en/t)^t$ and $1-p\leq e^{-p}$, the logarithm of the independent-set term divided by $t$ is at most
$$
\log(en/t)-\frac{p(t-1)}2\leq\frac12\log t-\frac32\log\log t+O(1)-2\log t+o(1),
$$
which tends to $-\infty$. That term is eventually below $1$. Thus $R(3,t)>5n/6-1$ and
$$
\boxed{R(3,t)=\Omega\!\left((t/\log t)^{3/2}\right).}
$$
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