If 1 man drinks 2L of water then 50 men would drink (on average) 100L of water. 0.7L standard deviation is equivalent to 35L for 50 men.
So if we assume a normal distribution with X=110 and μ=100, σ=35:
Z=(X-μ)/σ=(110-100)/35=10/35=0.29 approx.
This is equivalent to 0.6125, that is, the probability of using 110L of water or less is 61.25%.
Therefore there is a probability of 100-61.25=38.75% of running out of water.
You could also work it out by dividing 110L by 50=2.2L per man, μ=2, σ=0.7.
Z=(2.2-2.0)/0.7=0.2/0.7=0.29.