if $L\in BPP$ then $NP\subseteq BPP$
This is similar to the proof that PP contains NP. Given a 3SAT instance $\varphi$ on $n$ variables, construct a machine $M$ that with probability $3/4 - 2^{-n}/4$ accepts, and with probability $1/4 + 2^{-n}/4$ chooses a random truth assignment and accepts if the truth assignment satisfies $\varphi$. If $\varphi$ has $N$ satisfying assignments, then $M$ accepts with probability
$$
p = \frac{3}{4} - \frac{2^{-n}}{4} + \frac{M}{2^n} \left(\frac{1}{4} + \frac{2^{-n}}{4}\right).
$$
If $M = 0$ then $p &lt; 3/4$, whereas if $M \geq 1$ then $p \geq 3/4 + 2^{-2n}/4$.

I'll let you fill in the rest of the argument.