For the car to travel in a circular path, a centripetal force is required. This force is provided by the static friction, $f_s$, between the tires and the road, directed towards the center of the curve.

Applying Newton's second law in the vertical direction, the normal force $N$ balances the gravitational force $mg$:

$$ \sum F_y = N - mg = 0 \implies N = mg $$

In the horizontal (radial) direction, the static friction provides the centripetal force, $F_c = ma_c$, where the centripetal acceleration is $a_c = \frac{v^2}{r}$:

$$ \sum F_r = f_s = m\frac{v^2}{r} $$

The static friction force has a maximum value, $f_{s, \text{max}} = \mu_s N$. The maximum speed, $v_{\text{max}}$, is achieved when the required centripetal force equals this maximum available static friction.

$$ f_{s, \text{max}} = m\frac{v_{\text{max}}^2}{r} $$

Substituting $f_{s, \text{max}} = \mu_s N$ and $N = mg$:

$$ \mu_s (mg) = m\frac{v_{\text{max}}^2}{r} $$

Solving for $v_{\text{max}}$, we find that the mass $m$ cancels out:

$$ \mu_s g = \frac{v_{\text{max}}^2}{r} $$ $$ v_{\text{max}} = \sqrt{\mu_s g r} $$

Now, we substitute the given values: $\mu_s = 0.3$, $r = 300$ m, and using $g = 9.8$ m/s$^2$.

$$ v_{\text{max}} = \sqrt{(0.3)(9.8 \text{ m/s}^2)(300 \text{ m})} $$ $$ v_{\text{max}} = \sqrt{882 \text{ m}^2/\text{s}^2} $$ $$ v_{\text{max}} \approx 29.7 \text{ m/s} $$