Assuming that about `20 MeV` of energy is released per fusion reaction
`._1H^2 +._1 H^2 rarr. _2He^3 + E +` other particles then the mass of `._1H^2` consumed per day in a fusion reactor of power `1` megawatt wil approximately be.

Explanation

P=1 MW =`10^6` W=`10^6 Js^(-1)`
t=1 day = 24 x 60 x 60 =86400 s
`therefore` Energy released in one day = Pt = `86400 xx 10^6` J
Energy released per fusion
=20 MeV
`=20xx1.6xx10^(-13)`
`=3.2xx10^(-12) J`
Mass of `._1H^2` consumed in one fusion `(._1H^2+._1H^2)`
=4u
`=4xx1.66xx10^(-27) kg`
`=6.64xx10^(-27)` kg
`=(6.64xx10^(-27))/(3.2xx10^(-12))xx86400xx10^6`
`=1.79xx10^(-4)` kg