The standard formula for the amount of moisture to remove during grain drying is \(W_r = W_1 \times \dfrac{M_1 - M_2}{100 - M_2}\), where \(W_1\) is the starting weight, \(M_1\) the harvesting moisture, and \(M_2\) the target safe-storage moisture, both measured on a wet basis.
Plugging in the numbers: \(W_r = 30{,}000 \times \dfrac{14-12}{100-12} = 30{,}000 \times \dfrac{2}{88} \approx 682\) kg.
That doesn't land exactly on any of the listed options, but 782 kg is closest in magnitude to this figure among the choices given (the other three options are off by whole orders of magnitude), which is why it is recorded as the answer here; readers checking the arithmetic closely should note the roughly 100 kg gap between the calculated and listed value.