To find the probability that out of 5 workmen, 3 or more will contract the disease, we can use the binomial probability formula. The binomial probability formula is used when there are exactly two mutually exclusive outcomes of a trial - 'success' and 'failure'. In this case, 'success' is contracting the disease, and 'failure' is not contracting the disease.

The binomial probability formula is given by:

$$ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} $$

Where:

$ P(X = k) $ is the probability of having exactly $ k $ successes in $ n $ trials.
$ \binom{n}{k} $ is the binomial coefficient, which equals $\frac{n!}{k!(n-k)!}$ and represents the number of ways to choose $ k $ successes from $ n $ trials.
$ p $ is the probability of success on a single trial.
$ (1-p) $ is the probability of failure on a single trial.

In this problem, $ p = 0.10 $ and $ n = 5 $. We want to calculate the probability that 3 or more workmen will contract the disease, meaning we seek $ P(X \geq 3) $. This is the sum of the probabilities of having exactly 3, exactly 4, and exactly 5 workmen contracting the disease, which is:

$$ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) $$

Calculating the probabilities for each of these:

$ P(X = 3) = \binom{5}{3} (0.10)^3 (0.90)^2 $
$ P(X = 4) = \binom{5}{4} (0.10)^4 (0.90)^1 $
$ P(X = 5) = \binom{5}{5} (0.10)^5 (0.90)^0 $

Now we calculate these probabilities:

$ P(X = 3) = \binom{5}{3} (0.10)^3 (0.90)^2 $

$ P(X = 3) = \frac{5!}{3!2!} \times 0.001 \times 0.81 $

$ P(X = 3) = 10 \times 0.001 \times 0.81 $

$ P(X = 3) = 0.0081 $
$ P(X = 4) = \binom{5}{4} (0.10)^4 (0.90)^1 $

$ P(X = 4) = \frac{5!}{4!1!} \times 0.0001 \times 0.9 $

$ P(X = 4) = 5 \times 0.0001 \times 0.9 $

$ P(X = 4) = 0.00045 $
$ P(X = 5) = \binom{5}{5} (0.10)^5 (0.90)^0 $

$ P(X = 5) = \frac{5!}{5!0!} \times 0.00001 \times 1 $

$ P(X = 5) = 1 \times 0.00001 \times 1 $

$ P(X = 5) = 0.00001 $

Adding these probabilities gives us:

$$ P(X \geq 3) = P(X = 3) + P(X = 4) + P(X = 5) = 0.0081 + 0.00045 + 0.00001 $$

$$ P(X \geq 3) = 0.00856 $$

The answer is Option D, 0.00856.

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