$\because E=h v=\frac{h c}{\lambda}=h c \cdot \bar{v} \quad\left(\because \frac{1}{\lambda}=\bar{v}\right)$
where $E=$ energy of photon
$c=$ velocity of photon (-light)
$\lambda=$ wavelength of photon
$h=$ plank's constant.
$\therefore$ For $(\mathrm{a})$
$E=6.63 \times 10^{-34} \times 3 \times 10^{4} / 300 \times 10^{-9}$
$E=1.98 \times 10^{-25} \cdot \mathrm{J} / 300 \times 10^{-9}$
$\therefore \quad E=\frac{1.98 \times 10^{-2} \mathrm{J}}{300 \times 10^{-9} \mathrm{m}}$
$(a) \rightarrow E=6.6 \times 10^{-19} \mathrm{J}$
For (b) $E=h v=6.63 \times 10^{-34} \times 3 \times 10^{4}$
$(b) \rightarrow E=1.98 \times 10^{-35} \mathrm{J}$
For (c) $E=h c \times v$
$\because \frac{1}{3}$
$$
E=6.63 \times 10^{-4} \times 3 \times 10^{*} \times 30 \times 10^{2}
$$
(c) $\Rightarrow E=5.96 \times 10^{\circ}$
For $(d) d \rightarrow E=6.62 \times 10^{-27} J$
Hence, highest energy for photon is in (a).