Electron gives off radiation when it passes from the higher energy level to the lower (from the orbit with greater quantum number to one with the lower quantum number). The frequency always has to satisfy $hnu=Delta E$ where $Delta E$ is the difference in the energies of mentioned orbits. This then yields
$$
nu = frac{Delta E}{h},
$$
$h$ being Planck’s constant.

The radius increases. From the formula $r_n=(5.29times10^{-11}text{m})n^2$ we see that the radius is directly proportional to $n^2$ so increasing $n$ also increases the radius.

The energy increases. Note that in the expression for the energy of $n$th orbit
$$
E_n=-(13.6text{ eV})frac{1}{n^2}
$$
the absolute value of the energy decreases with the increase in $n$ (since we divide by $n^2$), but because there is minus sign in front then the actual energy increases with the increase in $n$ (it is less and less negative).

a) The Bohr radius of the orbit n = 6 can be calculated as follows:

$$
begin{align*}
r_n &= (5.29 times 10^{-11} text{ m})n^2 \
r_6 &= (5.29 times 10^{-11} text{ m})6^2 \
&= 1.9 times 10^{-9} text{ m}
end{align*}
$$

The Bohr radius is $boxed{r_6 = 1.9 times 10^{-9} text{ m}}$

b) The energy at this orbit can be calculated as follows:

$$
begin{align*}
E_n &= -(13.6 text{ eV})dfrac{1}{n^2} \
E_6 &= -(13.6 text{ eV})dfrac{1}{6^2} \
&= -0.38 text{ eV}
end{align*}
$$

The energy at n = 6 is $boxed{-0.38 text{ eV}}$

a) $r_6 = 1.9 times 10^{-9}$ m

b) $E_6 = -0.38$ eV

To be able to determine the Bohr electron orbit with the radius of $2.56 times 10^{-8}$ m, we can calculate n as follows:

$$
begin{align*}
r_n &= (5.29 times 10^{-11} text{ m})n^2 \
n^2 &= dfrac{2.56 times 10^{-8} text{ m}}{5.29 times 10^{-11} text{ m}} \
n &= sqrt{dfrac{2.56 times 10^{-8} text{ m}}{5.29 times 10^{-11} text{ m}}} \
&= 22
end{align*}
$$

Hence, the Bohr orbit of the electron is $boxed{text{n = 22}}$

$$
n = 22
$$