The precise answer depends on the initial condition, let's assume you start the Brownian motion (with diffusion constant $D$) at time $t=0$ in some point $x_0\in(a,b)$, then you ask for the probability $P(t)$ that the particle has not crossed the point $x=b$ for all times up to time $t$. The point $x=b$ functions as an absorbing boundary for the Brownian motion. The survival probability is given by $$P(t)={\rm Erf}\,\left(\frac{b-x_0}{2\sqrt{Dt}}\right)$$ For large $t$ this decays as as $1/\sqrt t$, $$P(t)=\frac{b-x_0}{\sqrt{\pi Dt}}+{\rm order}(1/t)$$ so much slower than exponentially.

The precise answer depends on the initial condition, let's assume you start the Brownian motion (with diffusion constant $D$) at time $t=0$ in some point $x_0\in(a,b)$, then you ask for the probability $$P(t)=\mathbb P[ \max\limits_{0 \leq s \leq t} B(s) \in (a,b) ]$$ that the particle has not crossed the point $x=b$ for all times up to time $t$. The point $x=b$ functions as an absorbing boundary for the Brownian motion. The survival probability is given by $$P(t)={\rm Erf}\,\left(\frac{b-x_0}{2\sqrt{Dt}}\right)$$ For large $t$ this decays as as $1/\sqrt t$, $$P(t)=\frac{b-x_0}{\sqrt{\pi Dt}}+{\rm order}(1/t)$$ so much slower than exponentially.