To find the speed of light in the denser medium, we can use Snell's Law at the critical angle, where the angle of refraction is $$90^{\circ}$$. Snell's Law states:

$$n_1 \sin{\theta_1} = n_2 \sin{\theta_2}$$

where $$n_1$$ and $$n_2$$ are the indices of refraction for the denser and rarer media, respectively, and $$\theta_1$$ and $$\theta_2$$ are the angles of incidence and refraction, respectively.

In this case, we have:

$$n_1 \sin{45^{\circ}} = n_2 \sin{90^{\circ}}$$

The critical angle is given as $$45^{\circ}$$, and the speed of light in the rarer medium is given as $$3 \times 10^8 \mathrm{~m/s}$$. We can find the index of refraction of the rarer medium using the formula:

$$n = \frac{c}{v}$$

where $$c$$ is the speed of light in a vacuum ($$3 \times 10^8 \mathrm{~m/s}$$), and $$v$$ is the speed of light in the medium.

For the rarer medium, we have:

$$n_2 = \frac{3 \times 10^8 \mathrm{~m/s}}{3 \times 10^8 \mathrm{~m/s}} = 1$$

Now we can rewrite Snell's Law as:

$$n_1 \sin{45^{\circ}} = 1 \cdot 1$$

$$n_1 \sin{45^{\circ}} = 1$$

$$n_1 = \frac{1}{\sin{45^{\circ}}}$$

Since $$\sin{45^{\circ}} = \frac{1}{\sqrt{2}}$$, we have:

$$n_1 = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2}$$

Now, we can find the speed of light in the denser medium using the formula for the index of refraction:

$$v_1 = \frac{c}{n_1}$$

$$v_1 = \frac{3 \times 10^8 \mathrm{~m/s}}{\sqrt{2}}$$

$$v_1 = 2.12 \times 10^8 \mathrm{~m/s}$$

So, the speed of light in the denser medium is $$2.12 \times 10^8 \mathrm{~m/s}$$.