A green LED has a semiconductor with a band gap of 2.3 eV, and each photon that it gives out has…

Step 1. The wavelength. $\lambda = \dfrac{hc}{E_g} = \dfrac{1240\ \text{eV nm}}{2.3\ \text{eV}} = 539$ nm.

Step 2. The electrons. $\dfrac{I}{e} = \dfrac{0.020\ \text{A}}{1.602 \times 10^{-19}\ \text{C}} = 1.25 \times 10^{17}\ \text{s}^{-1}$.

Step 3. The photons. $0.35 \times 1.25 \times 10^{17}\ \text{s}^{-1} = 4.37 \times 10^{16}\ \text{s}^{-1}$.

Step 4. The optical power. One photon has $2.3\ \text{eV} \times 1.602 \times 10^{-19}\ \text{J/eV} = 3.68 \times 10^{-19}\ \text{J}$. So $P_{opt} = 4.37 \times 10^{16}\ \text{s}^{-1} \times 3.68 \times 10^{-19}\ \text{J} = 16.1$ mW.

Step 5. The electrical power. $P_{el} = V I = 2.6\ \text{V} \times 20\ \text{mA} = 52.0$ mW.

Step 6. The efficiency. $\dfrac{16.1\ \text{mW}}{52.0\ \text{mW}} = 0.310 = 31.0\ \%$.