Photon flux and the power to 2 LEDS
\$20~ µmoles/m^2/s = 1080 ~lux = 4.38~ W/m^2 \$  for emitted power, not applied power.

\$1~ lux = 1~ lumen /m^2\$  

Efficacy=lumens/W  for applied power

thus 1080 lumens/m^2 / ? lumens/W x ? m^2 = Watts input power.

So you need to find l/W , and W for each LED then effective area ( then factor * efficiency for luminaire efficiency since you dont utilize 100% of emitted light on outside edges. )  it is usually 160 deg for half power for a lambertion source.

This is one example.

High flux output Far Red 740nm LED
 12.8 umol/s output at 5.7W power dissipation
 Ultra-small foot print – 7.0mm x 7.0mm
 Surface mount ceramic package with integrated glass lens
 Low Thermal Resistance (2.8°C/W)    

$18 http://www.ledengin.com/files/products/LZ4/LZ4-00R308.pdf

A CPU like heatsink is needed.