$P(f\in[10,20]) = P(120 \leq S_t \leq 130) = P(S_t \leq 130) - P(S_t \leq 120)$
That is, the probability that the option is between 10 and twenty is the same that the stock is between 120 and 130. The probability that the stock is between 120 and 130 is the probability the stock is less than 130 minus the probability that it is less than 120. If you are a more visual person consider this picture:
If we know that $W_t$ follows a driftless-Wiener process that:
$W_t - W_0 \sim N(0, 15^2t) \rightarrow S_t \sim N(95, 15^2t)$
If $S_t \sim N(95, 15^2t)$ then $P(S_t \leq 130) = $ prob. that a normal random variable with mean 95 and std. deviation 15t which is just the cumulative distribution function (CDF) of that normal random variable.
Here, that CDF is $\Phi( (x-95) / (15 \sqrt(t)))$, where $\Phi$ is the CDF of the standard normal. So the answer is $$\Phi( (130-95) / (15 \sqrt t)) - \Phi( (120-95) / (15 \sqrt t))$$
I thought that this was a simple option payoff but this is actually a trickier problem because it is a look back option taking the maximum stock value from 0 to 2. The general idea with this more complex problem is we are studying the maximum of a Wiener process motion rather than the Wiener process itself. Stochastic Processes and Related Distributions by Daniel Herlemont, section 2 should be of some help. But I don't have a solution to this problem at this time.