The number of amplitude modulated broadcast stations that can be accommodated in a 300 kHz band width for the highest modulating frequency 15 kHz will be:

For an amplitude modulated (AM) signal, the first thing to recall is the standard bandwidth formula:

$$B_{\text{AM}} \;=\; 2\,f_m^{\text{(max)}}$$

This formula states that the total bandwidth required for one AM broadcast is twice the highest audio (modulating) frequency present in that transmission, because the modulation process produces an upper sideband at $$f_c + f_m$$ and a lower sideband at $$f_c - f_m$$, each extending up to the highest modulating frequency $$f_m^{\text{(max)}}$$.

Now the question specifies the highest modulating frequency as $$f_m^{\text{(max)}} = 15\text{ kHz}$$. Substituting this value in the formula, we obtain the bandwidth needed for a single AM station:

$$\begin{aligned} B_{\text{per station}} &= 2 \times f_m^{\text{(max)}} \\ &= 2 \times 15\text{ kHz} \\ &= 30\text{ kHz}. \end{aligned}$$

Next, we are told that the total available broadcast band is $$B_{\text{total}} = 300\text{ kHz}$$. The number of completely non-overlapping AM stations that can fit into this band is simply the ratio of the total band to the bandwidth per station:

$$\begin{aligned} N &= \frac{B_{\text{total}}}{B_{\text{per station}}} \\ &= \frac{300\text{ kHz}}{30\text{ kHz}} \\ &= 10. \end{aligned}$$

So, exactly ten amplitude-modulated broadcast stations can be accommodated within the given 300 kHz band.

Hence, the correct answer is Option B.