Solution:

Let $p_{i}$ be the probability that the die lands on the number $i$. The problem gives that $p_{1}^{2}=0.04$, $p_{6}^{2}=0.01$, so we have
$$p_{1}=0.2, \quad p_{6}=0.1, \quad p_{2}+p_{3}+p_{4}+p_{5}=0.7$$
We are asked to maximize
$$2\left(p_{1} p_{6}+p_{2} p_{5}+p_{3} p_{4}\right)=2\left(0.02+p_{2} p_{5}+p_{3} p_{4}\right)$$
Let $x=p_{2}+p_{5}$ and $y=p_{3}+p_{4}$. Then by AM-GM, $p_{2} p_{5} \leq \frac{x^{2}}{4}$, $p_{3} p_{4} \leq \frac{y^{2}}{4}$. Also,
$$\frac{x^{2}+y^{2}}{4} \leq \frac{x^{2}+2 x y+y^{2}}{4}=\frac{(x+y)^{2}}{4}=\frac{0.7^{2}}{4}=0.1225$$
Hence,
$$2\left(p_{1} p_{6}+p_{2} p_{5}+p_{3} p_{4}\right) \leq 2(0.02+0.1225)=0.285$$
where equality holds if $p_{2}=p_{5}=0.35$, $p_{3}=p_{4}=0$.

Thus, we conclude that $p=0.285$ and $\lfloor 100 p\rfloor=28$.