= Solution
A standard <Brownian motion> starts at zero, has continuous sample paths and independent increments, with $W_t-W_s\sim N(0,t-s)$ for $0\leq s<t$.
Assume $S_0,K,T>0$ and $\sigma>0$; changing the sign of $\sigma$ just changes the sign of Brownian motion. <Itô formula> gives $dS_t=S_t[(\mu+\sigma^2/2)dt+\sigma\,dW_t]$. In the complete <Black-Scholes model>, the <risk-neutral pricing> theorem prices an integrable payoff $Y$ by $e^{-rT}\mathbb E_QY$, where the discounted asset is a martingale. <Girsanov theorem> therefore replaces the logarithmic drift by $r-\sigma^2/2$:
$$
\log S_T=\log S_0+(r-\sigma^2/2)T+\sigma\sqrt T\,Z,\qquad Z\sim N(0,1)\text{ under }Q.
$$
Let $d_2=[\log(S_0/K)+(r-\sigma^2/2)T]/(\sigma\sqrt T)$ and $d_1=d_2+\sigma\sqrt T$. The exercise event is $Z>-d_2$. Completing the square in the Gaussian integral yields
$$
\mathbb E[e^{aZ}\mathbf1_{\{Z>b\}}]=e^{a^2/2}\Phi(a-b).
$$
Consequently $\mathbb P_Q(S_T>K)=\Phi(d_2)$ and $\mathbb E_Q[S_T\mathbf1_{\{S_T>K\}}]=S_0e^{rT}\Phi(d_1)$. Thus the <European call option> price is
$$
\boxed{C_0=S_0\Phi(d_1)-Ke^{-rT}\Phi(d_2).}
$$
The physical drift $\mu$ disappears because the price is obtained under the martingale measure. If $\sigma=0$ in an arbitrage-free deterministic model, $S_t=S_0e^{rt}$ and the formula is interpreted by its zero-volatility limit.
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