The area of a triangle $ABC$, right-angled at $B$, is $\left(7 + \sqrt{192}\right)\ \text{cm}^{2}$. $AB$ has a length of $\left(1 + \sqrt{3}\right)\ \text{cm}$.

1. Set up the equation for the area of the triangle.

$\frac{1}{2} \times AB \times BC = \text{Area}$
$\frac{1}{2}(1 + \sqrt{3})BC = 7 + \sqrt{192}$

2. Simplify the surd $\sqrt{192}$ and isolate $BC$.

$\sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3}$
$BC = \frac{2(7 + 8\sqrt{3})}{1 + \sqrt{3}}$
$BC = \frac{14 + 16\sqrt{3}}{1 + \sqrt{3}}$

3. Rationalise the denominator.

$BC = \frac{(14 + 16\sqrt{3})(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})}$

4. Expand and simplify to find the exact length in the required form.

$BC = \frac{14 - 14\sqrt{3} + 16\sqrt{3} - 16(3)}{1 - 3}$
$BC = \frac{14 + 2\sqrt{3} - 48}{-2}$
$BC = \frac{-34 + 2\sqrt{3}}{-2}$
$BC = 17 - \sqrt{3}$

1. Find the length of $BC$ from the area given in part (i)

Area $= 7 + \sqrt{192} = 7 + 8\sqrt{3}$
$\frac{1}{2} \times AB \times BC = 7 + 8\sqrt{3}$
$BC = \frac{14 + 16\sqrt{3}}{1 + \sqrt{3}}$
$BC = \frac{(14 + 16\sqrt{3})(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})}$
$BC = \frac{14 - 14\sqrt{3} + 16\sqrt{3} - 48}{1 - 3}$
$BC = \frac{-34 + 2\sqrt{3}}{-2} = 17 - \sqrt{3}$

2. Apply Pythagoras' theorem to find $(AC)^2$

$(AC)^2 = AB^2 + BC^2$
$(AC)^2 = (1 + \sqrt{3})^2 + (17 - \sqrt{3})^2$

3. Expand and simplify the surd expressions

$(AC)^2 = (1 + 2\sqrt{3} + 3) + (289 - 34\sqrt{3} + 3)$
$(AC)^2 = (4 + 2\sqrt{3}) + (292 - 34\sqrt{3})$
$(AC)^2 = 296 - 32\sqrt{3}$