 Solution

False. For every deterministic $t>0$, Brownian scaling gives $B_t/\sqrt{t}\sim N(0,1)$. In particular, for any $\epsilon >0$,independently of $t$. Almost-sure convergence would imply convergence in probability, but these fixed positive tail probabilities rule out convergence in probability to zero. The square-root normalization retains fluctuations of order one.